# Project Euler: Python solutions

I'm trying the solve Project Euler with Python. My target is < 10 seconds per problem. Some common functions are in these modules:
prime.py
``` 1 prime_list = [2, 3, 5, 7, 11, 13, 17, 19, 23]   # Ensure that this is initialised with at least 1 prime
2 prime_dict = dict.fromkeys(prime_list, 1)
3 lastn      = prime_list[-1]
4
5 def _isprime(n):
6     ''' Raw check to see if n is prime. Assumes that prime_list is already populated '''
7     isprime = n >= 2 and 1 or 0
8     for prime in prime_list:                    # Check for factors with all primes
9         if prime * prime > n: break             # ... up to sqrt(n)
10         if not n % prime:
11             isprime = 0
12             break
13     if isprime: prime_dict[n] = 1               # Maintain a dictionary for fast lookup
14     return isprime
15
16 def _refresh(x):
17     ''' Refreshes primes upto x '''
18     global lastn
19     while lastn <= x:                           # Keep working until we've got up to x
20         lastn = lastn + 1                       # Check the next number
21         if _isprime(lastn):
22             prime_list.append(lastn)            # Maintain a list for sequential access
23
24 def prime(x):
25     ''' Returns the xth prime '''
26     global lastn
27     while len(prime_list) <= x:                 # Keep working until we've got the xth prime
28         lastn = lastn + 1                       # Check the next number
29         if _isprime(lastn):
30             prime_list.append(lastn)            # Maintain a list for sequential access
31     return prime_list[x]
32
33 def isprime(x):
34     ''' Returns 1 if x is prime, 0 if not. Uses a pre-computed dictionary '''
35     _refresh(x)                                 # Compute primes up to x (which is a bit wasteful)
36     return prime_dict.get(x, 0)                 # Check if x is in the list
37
38 def factors(n):
39     ''' Returns a prime factors of n as a list '''
40     _refresh(n)
41     x, xp, f = 0, prime_list[0], []
42     while xp <= n:
43         if not n % xp:
44             f.append(xp)
45             n = n / xp
46         else:
47             x = x + 1
48             xp = prime_list[x]
49     return f
50
51 def all_factors(n):
52    ''' Returns all factors of n, including 1 and n '''
53    f = factors(n)
54    elts = sorted(set(f))
55    numelts = len(elts)
56    def gen_inner(i):
57        if i >= numelts:
58            yield 1
59            return
60        thiselt = elts[i]
61        thismax = f.count(thiselt)
62        powers = [1]
63        for j in xrange(thismax):
64            powers.append(powers[-1] * thiselt)
65        for d in gen_inner(i+1):
66            for prime_power in powers:
67                yield prime_power * d
68    for d in gen_inner(0):
69        yield d
70
71 def num_factors(n):
72     ''' Returns the number of factors of n, including 1 and n '''
73     div = 1
74     x = 0
75     while n > 1:
76         c = 1
77         while not n % prime(x):
78             c = c + 1
79             n = n / prime(x)
80         x = x + 1
81         div = div * c
82     return div
83
84
85 '''
86 Optimisation notes:
87 # Function access FAIRLY slower than Array access is SLIGHTLY slower than local variable access
88 # Generators are SLIGHTLY slower than returning a list
89 # Dictionary access takes ZERO time
90 '''
```

combinatorics.py
``` 1 ''' From O'Reilly's Python Cookbook '''
2
3 def _combinators(_handle, items, n):
4     if n==0:
5         yield []
6         return
7     for i, item in enumerate(items):
8         this_one = [ item ]
9         for cc in _combinators(_handle, _handle(items, i), n-1):
10             yield this_one + cc
11
12 def combinations(items, n):
13     ''' take n distinct items, order matters '''
14     def skipIthItem(items, i):
15         return items[:i] + items[i+1:]
16     return _combinators(skipIthItem, items, n)
17
18 def uniqueCombinations(items, n):
19     ''' take n distinct items, order is irrelevant '''
20     def afterIthItem(items, i):
21         return items[i+1:]
22     return _combinators(afterIthItem, items, n)
23
24 def selections(items, n):
25     ''' take n (not necessarily distinct) items, order matters '''
26     def keepAllItems(items, i):
27         return items
28     return _combinators(keepAllItems, items, n)
29
30 def permutations(items):
31     ''' take all items, order matters '''
32     return combinations(items, len(items))
```
1233168157Show solution
``` 1 '''
2 If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
3
4 Find the sum of all the multiples of 3 or 5 below 1000.
5 '''
6
7 n = 0
8 for i in xrange(1,1000):
9     if not i % 5 or not i % 3:
10         n = n + i
11
12 print n
```
24613732139Show solution
``` 1 '''
2 Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
3
4 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
5
6 Find the sum of all the even-valued terms in the sequence which do not exceed four million.
7 '''
8
9 cache = {}
10 def fib(n):
11     cache[n] = cache.get(n, 0) or (n <= 1 and 1 or fib(n-1) + fib(n-2))
12     return cache[n]
13
14 n = 0
15 i = 0
16 while fib(i) <= 4000000:
17     if not fib(i) % 2: n = n + fib(i)
18     i = i + 1
19
20 print n
```
36857141Show solution
``` 1 '''
2 The prime factors of 13195 are 5, 7, 13 and 29.
3
4 What is the largest prime factor of the number 600851475143 ?
5 '''
6
7 n = 600851475143
8 i = 2
9 while i * i < n:
10     while n % i == 0:
11         n = n / i
12     i = i + 1
13
14 print n
```
4906609531Show solution
``` 1 '''
2 A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 x 99.
3
4 Find the largest palindrome made from the product of two 3-digit numbers.
5 '''
6
7 n = 0
8 for a in xrange(999, 100, -1):
9     for b in xrange(a, 100, -1):
10         x = a * b
11         if x > n:
12             s = str(a * b)
13             if s == s[::-1]:
14                 n = a * b
15
16 print n
```
5232792560140Show solution
``` 1 '''
2 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
3
4 What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?
5 '''
6
7 def gcd(a,b): return b and gcd(b, a % b) or a
8 def lcm(a,b): return a * b / gcd(a,b)
9
10 n = 1
11 for i in xrange(1, 21):
12     n = lcm(n, i)
13
14 print n
```
625164150171Show solution
``` 1 '''
2 The sum of the squares of the first ten natural numbers is,
3 1^2 + 2^2 + ... + 10^2 = 385
4 The square of the sum of the first ten natural numbers is,
5 (1 + 2 + ... + 10)^2 = 552 = 3025
6 Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 - 385 = 2640.
7
8 Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
9 '''
10
11 r = xrange(1, 101)
12 a = sum(r)
13 print a * a - sum(i*i for i in r)
```
7104743812Show solution
```1 '''
2 By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
3
4 What is the 10001st prime number?
5 '''
6
7 import prime
8 print prime.prime(10000)
```
840824187Show solution
``` 1 '''
2 Find the greatest product of five consecutive digits in the 1000-digit number.
3
4 73167176531330624919225119674426574742355349194934
5 96983520312774506326239578318016984801869478851843
6 85861560789112949495459501737958331952853208805511
7 12540698747158523863050715693290963295227443043557
8 66896648950445244523161731856403098711121722383113
9 62229893423380308135336276614282806444486645238749
10 30358907296290491560440772390713810515859307960866
11 70172427121883998797908792274921901699720888093776
12 65727333001053367881220235421809751254540594752243
13 52584907711670556013604839586446706324415722155397
14 53697817977846174064955149290862569321978468622482
15 83972241375657056057490261407972968652414535100474
16 82166370484403199890008895243450658541227588666881
17 16427171479924442928230863465674813919123162824586
18 17866458359124566529476545682848912883142607690042
19 24219022671055626321111109370544217506941658960408
20 07198403850962455444362981230987879927244284909188
21 84580156166097919133875499200524063689912560717606
22 05886116467109405077541002256983155200055935729725
23 71636269561882670428252483600823257530420752963450
24 '''
25
26 s = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'
27 n = 0
28 for i in xrange(0, len(s)-4):
29     p = 1
30     for j in xrange(i,i+5):
31         p = p * int(s[j])
32     if p > n: n = p
33
34 print n
```
931875000735Show solution
``` 1 '''
2 A Pythagorean triplet is a set of three natural numbers, a  b  c, for which,
3 a^2 + b^2 = c^2
4 For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
5
6 There exists exactly one Pythagorean triplet for which a + b + c = 1000.
7 Find the product abc.
8 '''
9
10 for a in xrange(1, 1000):
11     for b in xrange(a, 1000):
12         c = 1000 - a - b
13         if c > 0:
14             if c*c == a*a + b*b:
15                 print a*b*c
16                 break
```
101429138289222578Show solution
``` 1 '''
2 The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
3
4 Find the sum of all the primes below two million.
5 '''
6
7 sieve = [True] * 2000000    # Sieve is faster for 2M primes
8
9 def mark(sieve, x):
10     for i in xrange(x+x, len(sieve), x):
11         sieve[i] = False
12
13 for x in xrange(2, int(len(sieve) ** 0.5) + 1):
14     if sieve[x]: mark(sieve, x)
15
16 print sum(i for i in xrange(2, len(sieve)) if sieve[i])
```
1170600674187Show solution
``` 1 '''
2 In the 2020 grid below, four numbers along a diagonal line have been marked in red.
3
4 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
5 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
6 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
7 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
8 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
9 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
10 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
11 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
12 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
13 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
14 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
15 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
16 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
17 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
18 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
19 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
20 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
21 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
22 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
23 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
24 The product of these numbers is 26 x 63 x 78 x 14 = 1788696.
25
26 What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid?
27 '''
28 nums = (
29     ( 8, 2,22,97,38,15, 0,40, 0,75, 4, 5, 7,78,52,12,50,77,91, 8,),
30     (49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48, 4,56,62, 0,),
31     (81,49,31,73,55,79,14,29,93,71,40,67,53,88,30, 3,49,13,36,65,),
32     (52,70,95,23, 4,60,11,42,69,24,68,56, 1,32,56,71,37, 2,36,91,),
33     (22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80,),
34     (24,47,32,60,99, 3,45, 2,44,75,33,53,78,36,84,20,35,17,12,50,),
35     (32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70,),
36     (67,26,20,68, 2,62,12,20,95,63,94,39,63, 8,40,91,66,49,94,21,),
37     (24,55,58, 5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72,),
38     (21,36,23, 9,75, 0,76,44,20,45,35,14, 0,61,33,97,34,31,33,95,),
39     (78,17,53,28,22,75,31,67,15,94, 3,80, 4,62,16,14, 9,53,56,92,),
40     (16,39, 5,42,96,35,31,47,55,58,88,24, 0,17,54,24,36,29,85,57,),
41     (86,56, 0,48,35,71,89, 7, 5,44,44,37,44,60,21,58,51,54,17,58,),
42     (19,80,81,68, 5,94,47,69,28,73,92,13,86,52,17,77, 4,89,55,40,),
43     ( 4,52, 8,83,97,35,99,16, 7,97,57,32,16,26,26,79,33,27,98,66,),
44     (88,36,68,87,57,62,20,72, 3,46,33,67,46,55,12,32,63,93,53,69,),
45     ( 4,42,16,73,38,25,39,11,24,94,72,18, 8,46,29,32,40,62,76,36,),
46     (20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74, 4,36,16,),
47     (20,73,35,29,78,31,90, 1,74,31,49,71,48,86,81,16,23,57, 5,54,),
48     ( 1,70,54,71,83,51,54,69,16,92,33,48,61,43,52, 1,89,19,67,48,),
49 )
50
51 def seqs(nums, row, col):
52     if row + 4 <= len(nums):                                yield list(nums[i][col] for i in xrange(row, row+4))
53     if col + 4 <= len(nums[row]):                           yield list(nums[row][i] for i in xrange(col, col+4))
54     if row + 4 <= len(nums) and col + 4 <= len(nums[row]):  yield list(nums[row+i][col+i] for i in xrange(0,4))
55     if row + 4 <= len(nums) and col >= 3:                   yield list(nums[row+i][col-i] for i in xrange(0,4))
56
57 def product(seq):
58     n = 1
59     for x in seq: n = n * x
60     return n
61
62 def list_seqs(nums):
63     for row in xrange(0, len(nums)):
64         for col in xrange(0, len(nums[row])):
65             for seq in seqs(nums, row, col):
66                 yield seq
67
68 print max(product(seq) for seq in list_seqs(nums))
```
12765765007593Show solution
``` 1 '''
2 The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
3
4 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
5
6 Let us list the factors of the first seven triangle numbers:
7
8  1: 1
9  3: 1,3
10  6: 1,2,3,6
11 10: 1,2,5,10
12 15: 1,3,5,15
13 21: 1,3,7,21
14 28: 1,2,4,7,14,28
15 We can see that 28 is the first triangle number to have over five divisors.
16
17 What is the value of the first triangle number to have over five hundred divisors?
18 '''
19
20 import prime
21
22 for i in xrange(1, 1000000000):
23     n = i * (i+1) / 2
24     if prime.num_factors(n) > 500:
25         print n
26         break
```
135537376230141Show solution
```  1 '''
2 Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
3
4 37107287533902102798797998220837590246510135740250
5 46376937677490009712648124896970078050417018260538
6 74324986199524741059474233309513058123726617309629
7 91942213363574161572522430563301811072406154908250
8 23067588207539346171171980310421047513778063246676
9 89261670696623633820136378418383684178734361726757
10 28112879812849979408065481931592621691275889832738
11 44274228917432520321923589422876796487670272189318
12 47451445736001306439091167216856844588711603153276
13 70386486105843025439939619828917593665686757934951
14 62176457141856560629502157223196586755079324193331
15 64906352462741904929101432445813822663347944758178
16 92575867718337217661963751590579239728245598838407
17 58203565325359399008402633568948830189458628227828
18 80181199384826282014278194139940567587151170094390
19 35398664372827112653829987240784473053190104293586
20 86515506006295864861532075273371959191420517255829
21 71693888707715466499115593487603532921714970056938
22 54370070576826684624621495650076471787294438377604
23 53282654108756828443191190634694037855217779295145
24 36123272525000296071075082563815656710885258350721
25 45876576172410976447339110607218265236877223636045
26 17423706905851860660448207621209813287860733969412
27 81142660418086830619328460811191061556940512689692
28 51934325451728388641918047049293215058642563049483
29 62467221648435076201727918039944693004732956340691
30 15732444386908125794514089057706229429197107928209
31 55037687525678773091862540744969844508330393682126
32 18336384825330154686196124348767681297534375946515
33 80386287592878490201521685554828717201219257766954
34 78182833757993103614740356856449095527097864797581
35 16726320100436897842553539920931837441497806860984
36 48403098129077791799088218795327364475675590848030
37 87086987551392711854517078544161852424320693150332
38 59959406895756536782107074926966537676326235447210
39 69793950679652694742597709739166693763042633987085
40 41052684708299085211399427365734116182760315001271
41 65378607361501080857009149939512557028198746004375
42 35829035317434717326932123578154982629742552737307
43 94953759765105305946966067683156574377167401875275
44 88902802571733229619176668713819931811048770190271
45 25267680276078003013678680992525463401061632866526
46 36270218540497705585629946580636237993140746255962
47 24074486908231174977792365466257246923322810917141
48 91430288197103288597806669760892938638285025333403
49 34413065578016127815921815005561868836468420090470
50 23053081172816430487623791969842487255036638784583
51 11487696932154902810424020138335124462181441773470
52 63783299490636259666498587618221225225512486764533
53 67720186971698544312419572409913959008952310058822
54 95548255300263520781532296796249481641953868218774
55 76085327132285723110424803456124867697064507995236
56 37774242535411291684276865538926205024910326572967
57 23701913275725675285653248258265463092207058596522
58 29798860272258331913126375147341994889534765745501
59 18495701454879288984856827726077713721403798879715
60 38298203783031473527721580348144513491373226651381
61 34829543829199918180278916522431027392251122869539
62 40957953066405232632538044100059654939159879593635
63 29746152185502371307642255121183693803580388584903
64 41698116222072977186158236678424689157993532961922
65 62467957194401269043877107275048102390895523597457
66 23189706772547915061505504953922979530901129967519
67 86188088225875314529584099251203829009407770775672
68 11306739708304724483816533873502340845647058077308
69 82959174767140363198008187129011875491310547126581
70 97623331044818386269515456334926366572897563400500
71 42846280183517070527831839425882145521227251250327
72 55121603546981200581762165212827652751691296897789
73 32238195734329339946437501907836945765883352399886
74 75506164965184775180738168837861091527357929701337
75 62177842752192623401942399639168044983993173312731
76 32924185707147349566916674687634660915035914677504
77 99518671430235219628894890102423325116913619626622
78 73267460800591547471830798392868535206946944540724
79 76841822524674417161514036427982273348055556214818
80 97142617910342598647204516893989422179826088076852
81 87783646182799346313767754307809363333018982642090
82 10848802521674670883215120185883543223812876952786
83 71329612474782464538636993009049310363619763878039
84 62184073572399794223406235393808339651327408011116
85 66627891981488087797941876876144230030984490851411
86 60661826293682836764744779239180335110989069790714
87 85786944089552990653640447425576083659976645795096
88 66024396409905389607120198219976047599490197230297
89 64913982680032973156037120041377903785566085089252
90 16730939319872750275468906903707539413042652315011
91 94809377245048795150954100921645863754710598436791
92 78639167021187492431995700641917969777599028300699
93 15368713711936614952811305876380278410754449733078
94 40789923115535562561142322423255033685442488917353
95 44889911501440648020369068063960672322193204149535
96 41503128880339536053299340368006977710650566631954
97 81234880673210146739058568557934581403627822703280
98 82616570773948327592232845941706525094512325230608
99 22918802058777319719839450180888072429661980811197
100 77158542502016545090413245809786882778948721859617
101 72107838435069186155435662884062257473692284509516
102 20849603980134001723930671666823555245252804609722
103 53503534226472524250874054075591789781264330331690
104 '''
105
106 print str(sum((
107     37107287533902102798797998220837590246510135740250,
108     46376937677490009712648124896970078050417018260538,
109     74324986199524741059474233309513058123726617309629,
110     91942213363574161572522430563301811072406154908250,
111     23067588207539346171171980310421047513778063246676,
112     89261670696623633820136378418383684178734361726757,
113     28112879812849979408065481931592621691275889832738,
114     44274228917432520321923589422876796487670272189318,
115     47451445736001306439091167216856844588711603153276,
116     70386486105843025439939619828917593665686757934951,
117     62176457141856560629502157223196586755079324193331,
118     64906352462741904929101432445813822663347944758178,
119     92575867718337217661963751590579239728245598838407,
120     58203565325359399008402633568948830189458628227828,
121     80181199384826282014278194139940567587151170094390,
122     35398664372827112653829987240784473053190104293586,
123     86515506006295864861532075273371959191420517255829,
124     71693888707715466499115593487603532921714970056938,
125     54370070576826684624621495650076471787294438377604,
126     53282654108756828443191190634694037855217779295145,
127     36123272525000296071075082563815656710885258350721,
128     45876576172410976447339110607218265236877223636045,
129     17423706905851860660448207621209813287860733969412,
130     81142660418086830619328460811191061556940512689692,
131     51934325451728388641918047049293215058642563049483,
132     62467221648435076201727918039944693004732956340691,
133     15732444386908125794514089057706229429197107928209,
134     55037687525678773091862540744969844508330393682126,
135     18336384825330154686196124348767681297534375946515,
136     80386287592878490201521685554828717201219257766954,
137     78182833757993103614740356856449095527097864797581,
138     16726320100436897842553539920931837441497806860984,
139     48403098129077791799088218795327364475675590848030,
140     87086987551392711854517078544161852424320693150332,
141     59959406895756536782107074926966537676326235447210,
142     69793950679652694742597709739166693763042633987085,
143     41052684708299085211399427365734116182760315001271,
144     65378607361501080857009149939512557028198746004375,
145     35829035317434717326932123578154982629742552737307,
146     94953759765105305946966067683156574377167401875275,
147     88902802571733229619176668713819931811048770190271,
148     25267680276078003013678680992525463401061632866526,
149     36270218540497705585629946580636237993140746255962,
150     24074486908231174977792365466257246923322810917141,
151     91430288197103288597806669760892938638285025333403,
152     34413065578016127815921815005561868836468420090470,
153     23053081172816430487623791969842487255036638784583,
154     11487696932154902810424020138335124462181441773470,
155     63783299490636259666498587618221225225512486764533,
156     67720186971698544312419572409913959008952310058822,
157     95548255300263520781532296796249481641953868218774,
158     76085327132285723110424803456124867697064507995236,
159     37774242535411291684276865538926205024910326572967,
160     23701913275725675285653248258265463092207058596522,
161     29798860272258331913126375147341994889534765745501,
162     18495701454879288984856827726077713721403798879715,
163     38298203783031473527721580348144513491373226651381,
164     34829543829199918180278916522431027392251122869539,
165     40957953066405232632538044100059654939159879593635,
166     29746152185502371307642255121183693803580388584903,
167     41698116222072977186158236678424689157993532961922,
168     62467957194401269043877107275048102390895523597457,
169     23189706772547915061505504953922979530901129967519,
170     86188088225875314529584099251203829009407770775672,
171     11306739708304724483816533873502340845647058077308,
172     82959174767140363198008187129011875491310547126581,
173     97623331044818386269515456334926366572897563400500,
174     42846280183517070527831839425882145521227251250327,
175     55121603546981200581762165212827652751691296897789,
176     32238195734329339946437501907836945765883352399886,
177     75506164965184775180738168837861091527357929701337,
178     62177842752192623401942399639168044983993173312731,
179     32924185707147349566916674687634660915035914677504,
180     99518671430235219628894890102423325116913619626622,
181     73267460800591547471830798392868535206946944540724,
182     76841822524674417161514036427982273348055556214818,
183     97142617910342598647204516893989422179826088076852,
184     87783646182799346313767754307809363333018982642090,
185     10848802521674670883215120185883543223812876952786,
186     71329612474782464538636993009049310363619763878039,
187     62184073572399794223406235393808339651327408011116,
188     66627891981488087797941876876144230030984490851411,
189     60661826293682836764744779239180335110989069790714,
190     85786944089552990653640447425576083659976645795096,
191     66024396409905389607120198219976047599490197230297,
192     64913982680032973156037120041377903785566085089252,
193     16730939319872750275468906903707539413042652315011,
194     94809377245048795150954100921645863754710598436791,
195     78639167021187492431995700641917969777599028300699,
196     15368713711936614952811305876380278410754449733078,
197     40789923115535562561142322423255033685442488917353,
198     44889911501440648020369068063960672322193204149535,
199     41503128880339536053299340368006977710650566631954,
200     81234880673210146739058568557934581403627822703280,
201     82616570773948327592232845941706525094512325230608,
202     22918802058777319719839450180888072429661980811197,
203     77158542502016545090413245809786882778948721859617,
204     72107838435069186155435662884062257473692284509516,
205     20849603980134001723930671666823555245252804609722,
206     53503534226472524250874054075591789781264330331690,
207 )))[0:10]
```
148377999092Show solution
``` 1 '''
2 The following iterative sequence is defined for the set of positive integers
3 n -> n/2 (n is even)
4 n -> 3n + 1 (n is odd)
5
6 Using the rule above and starting with 13, we generate the following sequence:
7 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
8
9 It can be seen that this sequence (starting at 13 and fiinishing at 1) contains 10 terms.
10 Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
11
12 Which starting number, under one million, produces the longest chain?
13
14 NOTE: Once the chain starts the terms are allowed to go above one million.
15 '''
16
17 cache = { 1: 1 }
18 def chain(cache, n):
19     if not cache.get(n,0):
20         if n % 2: cache[n] = 1 + chain(cache, 3*n + 1)
21         else: cache[n] = 1 + chain(cache, n/2)
22     return cache[n]
23
24 m,n = 0,0
25 for i in xrange(1, 1000000):
26     c = chain(cache, i)
27     if c > m: m,n = c,i
28
29 print n
```
15137846528820141Show solution
``` 1 '''
2 Starting in the top left corner of a 2x2 grid, there are 6 routes (without backtracking) to the bottom right corner.
3
4 How many routes are there through a 20x20 grid?
5 '''
6
7 def fact(n):
8     f = 1
9     for x in xrange(1, n+1): f = f * x
10     return f
11
12 print fact(40) / fact(20) / fact(20)
```
161366141Show solution
``` 1 '''
2 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26
3 What is the sum of the digits of the number 2^1000?
4 '''
5
6 def digits(n):
7     s = 0
8     while n > 0:
9         s = s + (n % 10)
10         n = n / 10
11     return s
12
13 print digits(pow(2,1000))
```
1721124203Show solution
``` 1 '''
2 How many letters would be needed to write all the numbers in words from 1 to 1000?
3 '''
4
5 words = [
6     (   1,  'one'      , ''     ),
7     (   2,  'two'      , ''     ),
8     (   3,  'three'    , ''     ),
9     (   4,  'four'     , ''     ),
10     (   5,  'five'     , ''     ),
11     (   6,  'six'      , ''     ),
12     (   7,  'seven'    , ''     ),
13     (   8,  'eight'    , ''     ),
14     (   9,  'nine'     , ''     ),
15     (  10,  'ten'      , ''     ),
16     (  11,  'eleven'   , ''     ),
17     (  12,  'twelve'   , ''     ),
18     (  13,  'thirteen' , ''     ),
19     (  14,  'fourteen' , ''     ),
20     (  15,  'fifteen'  , ''     ),
21     (  16,  'sixteen'  , ''     ),
22     (  17,  'seventeen', ''     ),
23     (  18,  'eighteen' , ''     ),
24     (  19,  'nineteen' , ''     ),
25     (  20,  'twenty'   , ''     ),
26     (  30,  'thirty'   , ''     ),
27     (  40,  'forty'    , ''     ),
28     (  50,  'fifty'    , ''     ),
29     (  60,  'sixty'    , ''     ),
30     (  70,  'seventy'  , ''     ),
31     (  80,  'eighty'   , ''     ),
32     (  90,  'ninety'   , ''     ),
33     ( 100,  'hundred'  , 'and'  ),
34     (1000,  'thousand' , 'and'  ),
35 ]
36 words.reverse()
37
38 def spell(n, words):
39     word = []
40     while n > 0:
41         for num in words:
42             if num[0] <= n:
43                 div = n / num[0]
44                 n = n % num[0]
45                 if num[2]: word.append(' '.join(spell(div, words)))
46                 word.append(num[1])
47                 if num[2] and n: word.append(num[2])
48                 break
49     return word
50
51 print sum(len(word) for n in xrange(1, 1001) for word in spell(n, words))
```
181074437Show solution
``` 1 '''
2 By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
4 3
5 7 5
6 2 4 6
7 8 5 9 3
8
9 That is, 3 + 7 + 4 + 9 = 23.
10
11 Find the maximum total from top to bottom of the triangle below:
12
13 75
14 95 64
15 17 47 82
16 18 35 87 10
17 20 04 82 47 65
18 19 01 23 75 03 34
19 88 02 77 73 07 63 67
20 99 65 04 28 06 16 70 92
21 41 41 26 56 83 40 80 70 33
22 41 48 72 33 47 32 37 16 94 29
23 53 71 44 65 25 43 91 52 97 51 14
24 70 11 33 28 77 73 17 78 39 68 17 57
25 91 71 52 38 17 14 91 43 58 50 27 29 48
26 63 66 04 68 89 53 67 30 73 16 69 87 40 31
27 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
28
29 NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
30 '''
31
32 triangle = (
33     (75,                                                         ),
34     (95, 64,                                                     ),
35     (17, 47, 82,                                                 ),
36     (18, 35, 87, 10,                                             ),
37     (20,  4, 82, 47, 65,                                         ),
38     (19,  1, 23, 75,  3, 34,                                     ),
39     (88,  2, 77, 73,  7, 63, 67,                                 ),
40     (99, 65,  4, 28,  6, 16, 70, 92,                             ),
41     (41, 41, 26, 56, 83, 40, 80, 70, 33,                         ),
42     (41, 48, 72, 33, 47, 32, 37, 16, 94, 29,                     ),
43     (53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14,                 ),
44     (70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57,             ),
45     (91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48,         ),
46     (63, 66,  4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31,     ),
47     ( 4, 62, 98, 27, 23,  9, 70, 98, 73, 93, 38, 53, 60,  4, 23, ),
48 )
49
50 def path(triangle, num):
51     s = triangle[0][0]
52     col = 0
53     for row in xrange(1, len(triangle)):
54         if num % 2: col = col + 1
55         num = num / 2
56         s = s + triangle[row][col]
57     return s
58
59 print max(path(triangle, n) for n in xrange(0, 16384))
```
19171171Show solution
``` 1 '''
2 You are given the following information, but you may prefer to do some research for yourself.
3
4 1 Jan 1900 was a Monday.
5 Thirty days has September,
6 April, June and November.
7 All the rest have thirty-one,
8 Saving February alone,
9 Which has twenty-eight, rain or shine.
10 And on leap years, twenty-nine.
11 A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
12
13 How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
14 '''
15 import datetime
16
17 sundays = 0
18 for year in xrange(1901, 2001):
19     for month in xrange(1, 13):
20         d = datetime.date(year, month, 1)
21         if d.weekday() == 6:
22             sundays = sundays + 1
23
24 print sundays
```
20648140Show solution
``` 1 '''
2 Find the sum of digits in 100!
3 '''
4
5 def digits(n):
6     s = 0
7     while n > 0:
8         s = s + (n % 10)
9         n = n / 10
10     return s
11
12 n = 1
13 for i in xrange(1,100): n = n*i
14 print digits(n)
```
21316268625Show solution
``` 1 '''
2 Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
3 If d(a) = b and d(b) = a, where a  b, then a and b are an amicable pair and each of a and b are called amicable numbers.
4
5 For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
6
7 Evaluate the sum of all the amicable numbers under 10000.
8 '''
9
10 def divisors(n): return list(i for i in xrange(1, n/2+1) if n % i == 0)
11 pair = dict( ((n, sum(divisors(n))) for n in xrange(1, 10000)) )
12 print sum(n for n in xrange(1, 10000) if pair.get(pair[n], 0) == n and pair[n] != n)
```
22871198282203Show solution
``` 1 '''
2 Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.
3
4 For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 x 53 = 49714.
5
6 What is the total of all the name scores in the file?
7 '''
8
9 def worth(name): return sum(ord(letter) - ord('A') + 1 for letter in name)
10
12 names.sort()
13
14 print sum((i+1) * worth(names[i]) for i in xrange(0, len(names)))
```
23417987115281Show solution
``` 1 '''
2 A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
3
4 A number whose proper divisors are less than the number is called deficient and a number whose proper divisors exceed the number is called abundant.
5
6 As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
7
8 Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
9 '''
10
11 import prime
12
13 MAX = 28124
14 prime._refresh(MAX/2)
15 abundants = [n for n in xrange(1, MAX) if sum(prime.all_factors(n)) > n+n]
16 abundants_dict = dict.fromkeys(abundants, 1)
17
18 total = 0
19 for n in xrange(1, MAX):
20     sum_of_abundants = 0
21     for a in abundants:
22         if a > n: break
23         if abundants_dict.get(n - a):
24             sum_of_abundants = 1
25             break
26     if not sum_of_abundants:
27         total = total + n
28
29 print total
```
242783915460157Show solution
``` 1 '''
2 A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
3
4 012   021   102   120   201   210
5
6 What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
7 '''
8
9 def fact(n):
10     f = 1
11     for x in xrange(1, n+1): f = f * x
12     return f
13
14 def permutation(orig_nums, n):
15     nums = list(orig_nums)
16     perm = []
17     while len(nums):
18         divider = fact(len(nums)-1)
19         pos = n / divider
20         n = n % divider
21         perm.append(nums[pos])
22         nums = nums[0:pos] + nums[pos+1:]
23     return perm
24
25 print ''.join(str(x) for x in permutation(range(0,10), 999999))
```
254782140Show solution
``` 1 '''
2 What is the first term in the Fibonacci sequence to contain 1000 digits
3 '''
4
5 import math
6 phi = (1 + pow(5, 0.5)) / 2
7 c = math.log10(5) / 2
8 logphi = math.log10(phi)
9 n = 1
10 while True:
11     if n * logphi - c >= 999:
12         print n
13         break
14     n = n + 1
```
269831421Show solution
``` 1 '''
2 Find the value of d < 1000 for which 1 / d contains the longest recurring cycle
3 '''
4
5 def cycle_length(n):
6     i = 1
7     if n % 2 == 0: return cycle_length(n / 2)
8     if n % 5 == 0: return cycle_length(n / 5)
9     while True:
10         if (pow(10, i) - 1) % n == 0: return i
11         else: i = i + 1
12
13 m = 0
14 n = 0
15 for d in xrange(1,1000):
16     c = cycle_length(d)
17     if c > m:
18         m = c
19         n = d
20
21 print n
```
27-5923111108Show solution
``` 1 '''
2 Euler published the remarkable quadratic formula:
3
4 n^2 + n + 41
5
6 It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41^2 + 41 + 41 is clearly divisible by 41.
7
8 Using computers, the incredible formula  n^2 - 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.
9
10 Considering quadratics of the form:
11
12 n^2 + an + b, where |a| <= 1000 and |b| <= 1000
13
14 where |n| is the modulus/absolute value of n
15 e.g. |11| = 11 and |4| = 4
16 Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.
17
18 '''
19
20 import prime
21
22 max_pair = (0,0,0)
23 for a in xrange(-999, 1000):
24     for b in xrange(max(2, 1-a), 1000): # b >= 2, a + b + 1 >= 2
25         n, count = 0, 0
26         while True:
27             v = n*n + a*n + b
28             prime._refresh(v)
29             if prime.isprime(v): count = count + 1
30             else: break
31             n = n + 1
32         if count > max_pair[2]:
33             max_pair = (a,b,count)
34
35 print max_pair[0] * max_pair[1]
```
28669171001141Show solution
``` 1 '''
2 Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
3  43                49
4     21 22 23 24 25
5     20  7  8  9 10
6     19  6  1  2 11
7     18  5  4  3 12
8     17 16 15 14 13
9  37                31
10                       57
11 1
12 6*4
13 19*4
14 39*4
15 69*4
16
17 It can be verified that the sum of both diagonals is 101.
18
19 What is the sum of both diagonals in a 1001 by 1001 spiral formed in the same way?
20 '''
21
22 diagonal = 1
23 start = 1
24 for width in xrange(3, 1002, 2):
25     increment = width - 1
26     count = increment * 4
27     diagonal = diagonal + start * 4 + increment * 10
28     start = start + count
29
30 print diagonal
```
299183233Show solution
``` 1 '''
2 How many distinct terms are in the sequence generated by ab for 2 <= a <= 100 and 2 <= b <= 100
3 '''
4
5 terms = {}
6 count = 0
7 for a in xrange(2,101):
8     for b in xrange(2,101):
9         c = pow(a,b)
10         if not terms.get(c, 0):
11             terms[c] = 1
12             count = count + 1
13
14 print count
```
304438391703Show solution
``` 1 '''
2 Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
3
4 1634 = 1^4 + 6^4 + 3^4 + 4^4
5 8208 = 8^4 + 2^4 + 0^4 + 8^4
6 9474 = 9^4 + 4^4 + 7^4 + 4^4
7 As 1 = 1^4 is not a sum it is not included.
8
9 The sum of these numbers is 1634 + 8208 + 9474 = 19316.
10
11 Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
12 '''
13
14 def power_of_digits(n, p):
15     s = 0
16     while n > 0:
17         d = n % 10
18         n = n / 10
19         s = s + pow(d, p)
20     return s
21
22
23 print sum(n for n in xrange(2, 200000) if power_of_digits(n, 5) == n)
```
3173682230828Show solution
``` 1 '''
2 In England the currency is made up of pound, P, and pence, p, and there are eight coins in general circulation:
3
4 1p, 2p, 5p, 10p, 20p, 50p, P1 (100p) and P2 (200p).
5 It is possible to make P2 in the following way:
6
7 1 P1 + 1 50p + 2 20p + 1 5p + 1 2p + 3 1p
8 How many different ways can P2 be made using any number of coins?
9 '''
10
11 coins = (1, 2, 5, 10, 20, 50, 100, 200)
12
13 def balance(pattern): return sum(coins[x]*pattern[x] for x in xrange(0, len(pattern)))
14
15 def gen(pattern, coinnum, num):
16     coin = coins[coinnum]
17     for p in xrange(0, num/coin + 1):
18         newpat = pattern[:coinnum] + (p,)
19         bal = balance(newpat)
20         if bal > num: return
21         elif bal == num: yield newpat
22         elif coinnum < len(coins)-1:
23             for pat in gen(newpat, coinnum+1, num):
24                 yield pat
25
26 print sum(1 for pat in gen((), 0, 200))
```
324522843656Show solution
``` 1 '''
2 The product 7254 is unusual, as the identity, 39 x 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
3
4 Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
5
6 HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
7 '''
8
9 from combinatorics import permutations
10
11 def num(l):
12     s = 0
13     for n in l: s = s * 10 + n
14     return s
15
16 product = {}
17 for perm in permutations(range(1,10)):
18     for cross in range(1,4):            # Number can't be more than 4 digits
19         for eq in range(cross+1, 6):    # Result can't be less than 4 digits
20             a = num(perm[0:cross])
21             b = num(perm[cross:eq])
22             c = num(perm[eq:9])
23             if a * b == c: product[c] = 1
24
25 print sum(p for p in product)
```
33100171Show solution
``` 1 '''
2 The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
3
4 We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
5
6 There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.
7
8 If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
9 '''
10
11 def fractions():
12     for numerator in map(str, xrange(10, 100)):
13         for denominator in map(str, xrange(int(numerator)+1, 100)):
14             if numerator == denominator: continue
15             if numerator[1] == denominator[1] and numerator[1] == '0': continue
16             if numerator[0] == denominator[0] and int(numerator) * int(denominator[1]) == int(denominator) * int(numerator[1]): yield(int(numerator), int(denominator))
17             if numerator[0] == denominator[1] and int(numerator) * int(denominator[0]) == int(denominator) * int(numerator[1]): yield(int(numerator), int(denominator))
18             if numerator[1] == denominator[1] and int(numerator) * int(denominator[0]) == int(denominator) * int(numerator[0]): yield(int(numerator), int(denominator))
19             if numerator[1] == denominator[0] and int(numerator) * int(denominator[1]) == int(denominator) * int(numerator[0]): yield(int(numerator), int(denominator))
20
21 def gcd(a,b): return b and gcd(b, a % b) or a
22
23 numerator = 1
24 denominator = 1
25 for frac in fractions():
26     numerator = numerator * frac[0]
27     denominator = denominator * frac[1]
28
29 g = gcd(numerator, denominator)
30 print denominator / g
```
3440730671Show solution
``` 1 '''
2 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
3
4 Find the sum of all numbers which are equal to the sum of the factorial of their digits.
5
6 Note: as 1! = 1 and 2! = 2 are not sums they are not included.
7 '''
8
9 fact = (1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880)
10
11 def sum_of_digits_factorial(n):
12     s = 0
13     while n > 0:
14         d = n % 10
15         s = s + fact[d]
16         n = n / 10
17     return s
18
19 print sum(n for n in xrange(10, 100000) if n == sum_of_digits_factorial(n))
```
35555406Show solution
``` 1 '''
2 The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
3
4 There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
5
6 How many circular primes are there below one million?
7 '''
8
9 sieve = [True] * 1000000
10 sieve[0] = sieve[1] = False
11
12 def mark(sieve, x):
13     for i in xrange(x+x, len(sieve), x):
14         sieve[i] = False
15
16 for x in xrange(2, int(len(sieve) ** 0.5) + 1):
17     mark(sieve, x)
18
19 def circular(n):
20     digits = []
21     while n > 0:
22         digits.insert(0, str(n % 10))
23         n = n / 10
24     for d in xrange(1, len(digits)):
25         yield int(''.join(digits[d:] + digits[0:d]))
26
27 count = 0
28 for n, p in enumerate(sieve):
29     if p:
30         iscircularprime = 1
31         for m in circular(n):
32             if not sieve[m]:
33                 iscircularprime = 0
34                 break
35         if iscircularprime:
36             count = count + 1
37
38 print count
```
3687218722233Show solution
``` 1 '''
2 The decimal number, 585 = 10010010012 (binary), is palindromic in both bases.
3
4 Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
5
6 (Please note that the palindromic number, in either base, may not include leading zeros.)
7 '''
8
9 def ispalindrome(n, base):
10     digits = []
11     reverse = []
12     while n > 0:
13         d = str(n % base)
14         digits.append(d)
15         reverse.insert(0, d)
16         n = n / base
17     return digits == reverse
18
19 print sum(n for n in xrange(1, 1000000) if ispalindrome(n, 10) and ispalindrome(n, 2))
```
37748317234Show solution
``` 1 '''
2 The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
3
4 Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
5
6 NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
7 '''
8
9 import math, prime
10
11 digits = range(0, 10)
12 prime_digits = (2, 3, 5, 7)
13
14 def num(l):
15     s = 0
16     for n in l: s = s * 10 + n
17     return s
18
19 def is_left_truncatable(l):
20     is_truncatable = 1
21     for size in xrange(1, len(l)+1):
22         n = num(l[:size])
23         prime._refresh(int(math.sqrt(n)))
24         if not prime._isprime(n):
25             is_truncatable = 0
26             break
27     return is_truncatable
28
29 def is_right_truncatable(l):
30     is_truncatable = 1
31     for size in xrange(0, len(l)):
32         n = num(l[size:])
33         prime._refresh(int(math.sqrt(n)))
34         if not prime._isprime(n):
35             is_truncatable = 0
36             break
37     return is_truncatable
38
39 def gen(result, number):
40     if len(number) > 6: return
41     number = list(number)
42     number.append('')
43     for digit in digits:
44         number[-1] = digit
45         if is_left_truncatable(number):
46             if is_right_truncatable(number) and len(number) > 1:
47                 result.append(num(number))
48             gen(result, number)
49
50 result = []
51 gen(result, [])
52 print sum(result)
```
38932718654359Show solution
``` 1 '''
2 Take the number 192 and multiply it by each of 1, 2, and 3:
3
4 192 x 1 = 192
5 192 x 2 = 384
6 192 x 3 = 576
7 By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
8
9 The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
10
11 What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
12 '''
13
14 def get_pandigital(n):
15     pandigital = ''
16     for x in xrange(1, 10):
17         pandigital += str(x * n)
18         if len(pandigital) >= 9: break
19     if len(pandigital) == 9 and sorted(dict.fromkeys(list(pandigital)).keys()) == list("123456789"): return pandigital
20     else: return ''
21
22 max = 0
23 for n in xrange(1, 10000):
24     p = get_pandigital(n)
25     if p and p > max: max = p
26
27 print max
```
3984022046Show solution
``` 1 '''
2 If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
3
4 {20,48,52}, {24,45,51}, {30,40,50}
5
6 For which value of p < 1000, is the number of solutions maximised?
7 '''
8
9 maxp, maxsol = 0, 0
10 for p in xrange(12, 1001, 2):
11     solutions = 0
12     # a < b < c. So a is at most 1/3 of p. b is between a and (p-a)/2
13     for a in xrange(1, p/3):
14         a2 = a*a
15         for b in xrange(a, (p-a)/2):
16             c = p - a - b
17             if a2 + b*b == c*c: solutions = solutions + 1
18     if solutions > maxsol: maxp, maxsol = p, solutions
19
20 print maxp
```
40210157Show solution
``` 1 '''
2 An irrational decimal fraction is created by concatenating the positive integers:
3
4 0.123456789101112131415161718192021...
5
6 It can be seen that the 12th digit of the fractional part is 1.
7
8 If dn represents the nth digit of the fractional part, find the value of the following expression.
9
10 d1 x d10 x d100 x d1000 x d10000 x d100000 x d1000000
11
12 0 digit < 1
13 1 digit < + 9 * 1           10
14 2 digit < + 90 * 2          190
15 3 digit < + 900 * 3         2890
16 4 digit < + 9000 * 4
17 5 digit < + 90000 * 5
18 '''
19
20 def digit_at(n):
21     digits = 1
22     n = n - 1
23     while True:
24         numbers = 9 * pow(10, digits-1) * digits
25         if n > numbers: n = n - numbers
26         else: break
27         digits = digits + 1
28     num = n / digits + pow(10, digits-1)
29     return int(str(num)[n % digits])
30
31 print digit_at(1) * digit_at(10) * digit_at(100) * digit_at(1000) * digit_at(10000) * digit_at(100000) * digit_at(1000000)
```
417652413155Show solution
``` 1 '''
2 We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
3
4 What is the largest n-digit pandigital prime that exists?
5 '''
6
7 import prime
8 from combinatorics import permutations
9
10 # Pan-digital primes are 4 or 7 digits. Others divisible by 3
11 prime._refresh(2766)    # sqrt(7654321)
12 for perm in permutations(range(7, 0, -1)):
13     num = 0
14     for n in perm: num = num * 10 + n
15     if prime._isprime(num):
16         print num
17         break
```
42162187Show solution
``` 1 '''
2 The nth term of the sequence of triangle numbers is given by, t_n = 1/2 x n(n+1); so the first ten triangle numbers are:
3
4 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
5
6 By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t_10. If the word value is a triangle number then we shall call the word a triangle word.
7
8 Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words?
9
10 '''
11 def worth(word): return sum(ord(letter) - ord('A') + 1 for letter in word)
12
14 triangle_numbers = dict.fromkeys(list(n*(n+1)/2 for n in xrange(1, 100)), 1)
15
16 print sum(1 for word in words if worth(word) in triangle_numbers)
```
431669533489010156Show solution
``` 1 '''
2 The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.
3
4 Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
5
6 d2 d3 d4 = 406 is divisible by 2
7 d3 d4 d5 = 063 is divisible by 3
8 d4 d5 d6 = 635 is divisible by 5
9 d5 d6 d7 = 357 is divisible by 7
10 d6 d7 d8 = 572 is divisible by 11
11 d7 d8 d9 = 728 is divisible by 13
12 d8 d9 d10= 289 is divisible by 17
13
14 Find the sum of all 0 to 9 pandigital numbers with this property.
15 '''
16
17 from combinatorics import permutations
18
19 def num(l):
20     s = 0
21     for n in l: s = s * 10 + n
22     return s
23
24 def subdiv(l, n): return num(l) % n == 0
25
26 total = 0
27 for perm in permutations((0,1,2,3,4,6,7,8,9)):
28     perm.insert(5, 5)               # d6 must be 5
29     if (subdiv(perm[7:10], 17) and
30         subdiv(perm[6:9],  13) and
31         subdiv(perm[5:8],  11) and
32         subdiv(perm[4:7],   7) and
33         subdiv(perm[3:6],   5) and
34         subdiv(perm[2:5],   3) and
35         subdiv(perm[1:4],   2)):
36             total += num(perm)
37
38 print total
```
4454826604046Show solution
``` 1 '''
2 Pentagonal numbers are generated by the formula, P_n=n(3n-1)/2. The first ten pentagonal numbers are:
3
4 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
5
6 It can be seen that P_4 + P_7 = 22 + 70 = 92 = P_8. However, their difference, 70 - 22 = 48, is not pentagonal.
7
8 Find the pair of pentagonal numbers, P_j and P_k, for which their sum and difference is pentagonal and D = |P_k - P_j| is minimised; what is the value of D?
9 '''
10
11 MAX = 2000
12 pent = [ n * (3*n - 1) / 2 for n in xrange(1, 2*MAX) ]
13 pdic = dict.fromkeys(pent)
14
15 def main2():
16     for j in xrange(0, MAX):
17         for k in xrange(j+1, 2*MAX-1):
18             p_j = pent[j]
19             p_k = pent[k]
20             p_sum = p_j + p_k
21             p_diff = p_k - p_j
22             if pdic.has_key(p_sum) and pdic.has_key(p_diff):
23                 return p_diff
24
25 print main2()
```
451533776805812Show solution
``` 1 '''
2 Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
3
4 Triangle        T_n=n(n+1)/2        1, 3, 6, 10, 15, ...
5 Pentagonal      P_n=n(3n-1)/2       1, 5, 12, 22, 35, ...
6 Hexagonal       H_n=n(2n-1)         1, 6, 15, 28, 45, ...
7
8 It can be verified that T_285 = P_165 = H_143 = 40755.
9
10 Find the next triangle number that is also pentagonal and hexagonal.
11 '''
12
13 MAX = 100000
14 triangle = [ n * (  n + 1) / 2 for n in xrange(0, MAX) ]
15 pentagon = [ n * (3*n - 1) / 2 for n in xrange(0, MAX) ]
16 hexagon  = [ n * (2*n - 1)     for n in xrange(0, MAX) ]
17 pentagon_dict = dict.fromkeys(pentagon, 1)
18 hexagon_dict  = dict.fromkeys(hexagon, 1)
19
20 for t in xrange(286, MAX):
21     v = triangle[t]
22     if pentagon_dict.has_key(v) and hexagon_dict.has_key(v):
23         print v
24         break
```
465777437Show solution
``` 1 '''
2 It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
3
4  9 =  7 + 2 x 1^2
5 15 =  7 + 2 x 2^2
6 21 =  3 + 2 x 3^2
7 25 =  7 + 2 x 3^2
8 27 = 19 + 2 x 2^2
9 33 = 31 + 2 x 1^2
10 It turns out that the conjecture was false.
11
12 What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
13 '''
14
15 import prime
16 MAX = 10000
17 squares = dict.fromkeys((x*x for x in xrange(1, MAX)), 1)
18 prime._refresh(MAX)
19
20 for x in xrange(35, MAX, 2):
21     if not prime.isprime(x):
22         is_goldbach = 0
23         for p in prime.prime_list[1:]:
24             if p >= x: break
25             if squares.has_key((x - p)/2):
26                 is_goldbach = 1
27                 break
28         if not is_goldbach:
29             print x
30             break
```
4713404390046Show solution
``` 1 '''
2 Find the first four consecutive integers to have four distinct prime factors
3 '''
4
5 import prime
6
7 def distinct_factors(n): return len(dict.fromkeys(prime.factors(n)).keys())
8
9 factors = [0, 1, distinct_factors(2), distinct_factors(3)]
10 while True:
11     if factors[-4::] == [4,4,4,4]: break
12     else: factors.append(distinct_factors(len(factors)))
13
14 print len(factors)-4
```
489110846700344Show solution
``` 1 '''
2 Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
3 '''
4
5 s = 0
6 mod = pow(10, 10)
7 for x in xrange(1, 1001):
8     s = s + pow(x, x)
9
10 print s % mod
```
49296962999629515Show solution
``` 1 '''
2 The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.
3
4 There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
5
6 What 12-digit number do you form by concatenating the three terms in this sequence?
7
8 '''
9
10 import prime
11 from combinatorics import permutations
12
13 prime._refresh(10000)
14 for num in xrange(1000, 10000):
15     if str(num).find('0') >= 0: continue
16
17     if prime.isprime(num):
18         prime_permutations = { num: 1 }
19         for x in permutations(list(str(num))):
20             next_num = int(''.join(x))
21             if prime.isprime(next_num):
22                 prime_permutations[next_num] = 1
23
24         primes = sorted(prime_permutations.keys())
25         for a in xrange(0, len(primes)):
26             if primes[a] == 1487: continue
27             for b in xrange(a+1, len(primes)):
28                 c = (primes[a] + primes[b]) / 2
29                 if prime_permutations.has_key(c):
30                     print str(primes[a]) + str(c) + str(primes[b])
31                     exit()
```
5099765110625Show solution
``` 1 '''
2 The prime 41, can be written as the sum of six consecutive primes:
3
4 41 = 2 + 3 + 5 + 7 + 11 + 13
5 This is the longest sum of consecutive primes that adds to a prime below one-hundred.
6
7 The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
8
9 Which prime, below one-million, can be written as the sum of the most consecutive primes?
10 '''
11 import prime
12
13 MAX = 5000
14 prime.prime(MAX)
15
16 def check_length(n, below):
17     maxprime = 0
18     for x in xrange(0, below):
19         total = sum(prime.prime_list[x:x+n])
20         if total > below: break
21         if prime.isprime(total): maxprime = total
22     return maxprime
23
24 for n in xrange(1000, 0, -1):
25     maxprime = check_length(n, 1000000)
26     if maxprime:
27         print maxprime
28         break
```
5112131334968Show solution
``` 1 '''
2 By replacing the 1st digit of *57, it turns out that six of the possible values: 157, 257, 457, 557, 757, and 857, are all prime.
3
4 By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.
5
6 Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.
7 '''
8
9 import prime
10 from combinatorics import uniqueCombinations
11
12 cache = {}
13 def prime_family_length(n, digits):
14     if cache.has_key((n, digits)): return cache[n, digits]
15
16     num, nums, count = list(str(n)), [], 0
17     if len(dict.fromkeys(num[d] for d in digits).keys()) > 1:
18         return cache.setdefault((n, digits), 0)                                # The digits must have the same number
19
20     for d in range(0 in digits and 1 or 0, 10):                                 # Ensure 0 is not the first digit
21         for x in digits: num[x] = str(d)
22         n = int(''.join(num))
23         if prime.isprime(n): count += 1
24         nums.append(n)
25     for n in nums: cache[n, digits] = count
26     return count
27
28 prime._refresh(100000)
29
30 n, max, max_count, combos = 10, 0, 0, {}
31 while max_count < 8:
32     p = prime.prime(n)
33     digits = range(0, len(str(p)))
34     for size in xrange(1, len(digits)):
35         patterns = combos.setdefault((len(digits), size),
36             tuple(tuple(sorted(p)) for p in uniqueCombinations(digits, size)))
37         for pat in patterns:
38             count = prime_family_length(p, pat)
39             if count > max_count: max, max_count = p, count
40     n += 1
41
42 print p
```
52142857640Show solution
``` 1 '''
2 It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
3
4 Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
5 '''
6
7 def multiples_have_same_digits(n):
8     digit_keys = dict.fromkeys(list(str(n)))
9     for x in xrange(2, 4):
10         for d in list(str(x * n)):
11             if not digit_keys.has_key(d): return False
12     return True
13
14 n = 0
15 while True:
16     n = n + 9                           # n must be a multiple of 9 for this to happen
17     if multiples_have_same_digits(n):
18         print n
19         break
```
534075280Show solution
``` 1 '''
2 There are exactly ten ways of selecting three from five, 12345:
3
4 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
5
6 In combinatorics, we use the notation, 5C3 = 10.
7
8 In general,
9
10 nCr = n! / r!(nr)! where r <= n, n! = n x (n-1)...x 3 x 2 x 1, and 0! = 1.
11
12 It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
13
14 How many, not necessarily distinct, values of  nCr, for 1 <= n <= 100, are greater than one-million?
15 '''
16
17 fact_c = { 0: 1, 1: 1 }
18 def fact(n): return fact_c.has_key(n) and fact_c[n] or fact_c.setdefault(n, n * fact(n-1))
19
20 count = 0
21 for n in xrange(1, 101):
22     for r in xrange(0, n):
23         ncr = fact(n) / fact(r) / fact(n-r)
24         if ncr > 1000000: count += 1
25 print count
```
54376860Show solution
```  1 '''
2 In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:
3
4 High Card: Highest value card.
5 One Pair: Two cards of the same value.
6 Two Pairs: Two different pairs.
7 Three of a Kind: Three cards of the same value.
8 Straight: All cards are consecutive values.
9 Flush: All cards of the same suit.
10 Full House: Three of a kind and a pair.
11 Four of a Kind: Four cards of the same value.
12 Straight Flush: All cards are consecutive values of same suit.
13 Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.
14 The cards are valued in the order:
15 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
16
17 If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives (see example 1 below). But if two ranks tie, for example, both players have a pair of queens, then highest cards in each hand are compared (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.
18
19 Consider the following five hands dealt to two players:
20
21 Hand        Player 1            Player 2          Winner
22 1       5H 5C 6S 7S KD      2C 3S 8S 8D TD      Player 2
23         Pair of Fives       Pair of Eights
24
25 2       5D 8C 9S JS AC      2C 5C 7D 8S QH      Player 1
26         Highest card Ace    Highest card Queen
27
28 3       2D 9C AS AH AC      3D 6D 7D TD QD      Player 2
29         Three Aces          Flush with Diamonds
30
31 4       4D 6S 9H QH QC      3D 6D 7H QD QS      Player 1
32         Pair of Queens      Pair of Queens
33         Highest card Nine   Highest card Seven
34
35 5       2H 2D 4C 4D 4S      3C 3D 3S 9S 9D      Player 1
36         Full House          Full House
37         With Three Fours    with Three Threes
38
39 The file, poker.txt, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards. You can assume that all hands are valid (no invalid characters or repeated cards), each player's hand is in no specific order, and in each hand there is a clear winner.
40
41 How many hands does Player 1 win?
42 '''
43
44 value = { '2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'T':10,'J':11,'Q':12,'K':13,'A':14 }
45 all_kinds = tuple(reversed(sorted(value.values())))
46 all_suits = list('DCSH')
47
48 def make_hand(cards):
49     hand = {}
50     for card in cards:
51         hand.setdefault(value[card[0]], {})[card[1]] = 1
52         hand.setdefault(card[1], {})[value[card[0]]] = 1
53     return hand
54
55 def get(hash, arr): return ((i, hash.get(i, {})) for i in arr)
56 def has(hash, arr): return not sum(1 for i in arr if i not in hash)
57
58 def rank(hand):
59     # Royal flush
60     for suit, kinds in get(hand, all_suits):
61         if has(kinds, tuple('TJQKA')):
62             return (9,)
63
64     # Straight flush
65     for suit, kinds in get(hand, all_suits):
66         kinds = sorted(kind for kind in kinds.keys())
67         if len(kinds) == 5 and kinds[4] - kinds[0] == 4:
68             return (8, kinds[0])
69
70     # Four of a kind
71     for kind, suits in get(hand, all_kinds):
72         if len(suits.keys()) == 4:
73             return (7, kind)
74
75     # Full house
76     for kind, suits in get(hand, all_kinds):
77         if len(suits.keys()) == 3:
78             for kind2, suits2 in get(hand, all_kinds):
79                 if len(suits2.keys()) == 2:
80                     return (6, kind, kind2)
81
82     # Flush
83     for suit, kinds in get(hand, all_suits):
84         if len(kinds.keys()) == 5:
85             return (5,)
86
87     # Straight
88     kinds = sorted(kind for kind in all_kinds if hand.has_key(kind))
89     if len(kinds) == 5 and kinds[4] - kinds[0] == 4:
90         return (4, kinds[0])
91
92     # Three of a kind
93     for kind, suits in get(hand, all_kinds):
94         if len(suits.keys()) == 3:
95             return (3, kind)
96
97     # Two pairs
98     for kind, suits in get(hand, all_kinds):
99         if len(suits.keys()) == 2:
100             for kind2, suits2 in get(hand, all_kinds):
101                 if kind != kind2 and len(suits2.keys()) == 2:
102                     return (2, kind, kind2)
103
104     # One pair
105     for kind, suits in get(hand, all_kinds):
106         if len(suits.keys()) == 2:
107             return (1, kind)
108
109     for kind in all_kinds:
110         if kind in hand:
111             return (0, kind)
112
113     return 0
114
115
116 count = 0
117 for hand in open('poker.txt'):
118     hands = hand.split(' ')
119     p1, p2 = make_hand(hands[0:5]), make_hand(hands[5:10])
120     v1, v2 = rank(p1), rank(p2)
121     if v1 > v2: count += 1
122 print count
```
55249546Show solution
``` 1 '''
2 If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
3
4 Not all numbers produce palindromes so quickly. For example,
5
6 349 + 943 = 1292,
7 1292 + 2921 = 4213
8 4213 + 3124 = 7337
9
10 That is, 349 took three iterations to arrive at a palindrome.
11
12 Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).
13
14 Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.
15
16 How many Lychrel numbers are there below ten-thousand?
17
18 NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.
19 '''
20
21 def is_lychrel(n):
22     n = str(n)
23     for count in xrange(0, 50):
24         n = str(int(n) + int(n[::-1]))
25         if n == n[::-1]: return False
26     return True
27
28 print sum(1 for n in xrange(0, 10000) if is_lychrel(n))
```
569721922Show solution
``` 1 '''
2 A googol (10^100) is a massive number: one followed by one-hundred zeros; 100^100 is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.
3
4 Considering natural numbers of the form, a^b, where a, b < 100, what is the maximum digital sum?
5 '''
6
7 max = 0
8 for a in xrange(0, 100):
9     for b in xrange(0, 100):
10         ds = sum(int(digit) for digit in str(a**b))
11         if ds > max: max = ds
12 print max
```
57153328Show solution
``` 1 '''
2 It is possible to show that the square root of two can be expressed as an infinite continued fraction.
3
4 sqrt(2) = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
5
6 By expanding this for the first four iterations, we get:
7
8 1 + 1/2 = 3/2 = 1.5
9 1 + 1/(2 + 1/2) = 7/5 = 1.4
10 1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
11 1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
12
13 The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
14
15 In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
16 '''
17
18 num, den, count = 3, 2, 0
19 for iter in xrange(0, 1000):
20     num, den = num + den + den, num + den
21     if len(str(num)) > len(str(den)):
22         count += 1
23 print count
```
58262416952Show solution
``` 1 '''
2 Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
3
4 37 36 35 34 33 32 31
5 38 17 16 15 14 13 30
6 39 18  5  4  3 12 29
7 40 19  6  1  2 11 28
8 41 20  7  8  9 10 27
9 42 21 22 23 24 25 26
10 43 44 45 46 47 48 49
11
12 It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ~ 62%.
13
14 If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
15 '''
16
17 import prime
18 prime._refresh(50000)
19
20 width, diagonal, base, primes = 1, 1, 1, 0
21 while True:
22     width = width + 2
23     increment = width - 1
24     for i in xrange(0, 4):
25         diagonal = diagonal + increment
26         if i < 3 and prime._isprime(diagonal): primes += 1
27     base = base + 4
28     if primes * 10 < base:
29         print width
30         break
```
591073599203Show solution
``` 1 '''
2 Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
3
4 A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.
5
6 For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both "halves", it is impossible to decrypt the message.
7
8 Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.
9
10 Your task has been made easy, as the encryption key consists of three lower case characters. Using cipher1.txt (right click and 'Save Link/Target As...'), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.
11 '''
12
13 from combinatorics import selections
14
15 code = tuple(int(c) for c in open('cipher1.txt').read().split(','))
16