| 1 | 233168 | 139 | Show solution'''
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.
Find the sum of all the multiples of 3 or 5 below 1000.
'''
n = 0
for i in xrange(1,1000):
if not i % 5 or not i % 3:
n = n + i
print n
|
| 2 | 4613732 | 125 | Show solution'''
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:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
Find the sum of all the even-valued terms in the sequence which do not exceed four million.
'''
cache = {}
def fib(n):
cache[n] = cache.get(n, 0) or (n <= 1 and 1 or fib(n-1) + fib(n-2))
return cache[n]
n,i = 0,0
while fib(i) <= 4000000:
if not fib(i) % 2: n = n + fib(i)
i = i + 1
print n
|
| 3 | 6857 | 141 | Show solution'''
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
'''
n = 600851475143
i = 2
while i * i < n:
while n % i == 0:
n = n / i
i = i + 1
print n
|
| 4 | 906609 | 531 | Show solution'''
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.
Find the largest palindrome made from the product of two 3-digit numbers.
'''
n = 0
for a in xrange(999, 100, -1):
for b in xrange(a, 100, -1):
x = a * b
if x > n:
s = str(a * b)
if s == s[::-1]:
n = a * b
print n
|
| 5 | 232792560 | 140 | Show solution'''
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?
'''
def gcd(a,b): return b and gcd(b, a % b) or a
def lcm(a,b): return a * b / gcd(a,b)
n = 1
for i in xrange(1, 21):
n = lcm(n, i)
print n
|
| 6 | 25164150 | 171 | Show solution'''
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 552 = 3025
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.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
'''
r = xrange(1, 101)
a = sum(r)
print a * a - sum(i*i for i in r)
|
| 7 | 104743 | 812 | Show solution'''
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10001st prime number?
'''
import prime
print prime.prime(10000)
|
| 8 | 40824 | 187 | Show solution'''
Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
'''
s = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'
n = 0
for i in xrange(0, len(s)-4):
p = 1
for j in xrange(i,i+5):
p = p * int(s[j])
if p > n: n = p
print n
|
| 9 | 31875000 | 735 | Show solution'''
A Pythagorean triplet is a set of three natural numbers, a b c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
'''
for a in xrange(1, 1000):
for b in xrange(a, 1000):
c = 1000 - a - b
if c > 0:
if c*c == a*a + b*b:
print a*b*c
break
|
| 10 | 142913828922 | 2578 | Show solution'''
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
'''
sieve = [True] * 2000000 # Sieve is faster for 2M primes
def mark(sieve, x):
for i in xrange(x+x, len(sieve), x):
sieve[i] = False
for x in xrange(2, int(len(sieve) ** 0.5) + 1):
if sieve[x]: mark(sieve, x)
print sum(i for i in xrange(2, len(sieve)) if sieve[i])
|
| 11 | 70600674 | 187 | Show solution'''
In the 2020 grid below, four numbers along a diagonal line have been marked in red.
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
The product of these numbers is 26 x 63 x 78 x 14 = 1788696.
What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid?
'''
nums = (
( 8, 2,22,97,38,15, 0,40, 0,75, 4, 5, 7,78,52,12,50,77,91, 8,),
(49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48, 4,56,62, 0,),
(81,49,31,73,55,79,14,29,93,71,40,67,53,88,30, 3,49,13,36,65,),
(52,70,95,23, 4,60,11,42,69,24,68,56, 1,32,56,71,37, 2,36,91,),
(22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80,),
(24,47,32,60,99, 3,45, 2,44,75,33,53,78,36,84,20,35,17,12,50,),
(32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70,),
(67,26,20,68, 2,62,12,20,95,63,94,39,63, 8,40,91,66,49,94,21,),
(24,55,58, 5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72,),
(21,36,23, 9,75, 0,76,44,20,45,35,14, 0,61,33,97,34,31,33,95,),
(78,17,53,28,22,75,31,67,15,94, 3,80, 4,62,16,14, 9,53,56,92,),
(16,39, 5,42,96,35,31,47,55,58,88,24, 0,17,54,24,36,29,85,57,),
(86,56, 0,48,35,71,89, 7, 5,44,44,37,44,60,21,58,51,54,17,58,),
(19,80,81,68, 5,94,47,69,28,73,92,13,86,52,17,77, 4,89,55,40,),
( 4,52, 8,83,97,35,99,16, 7,97,57,32,16,26,26,79,33,27,98,66,),
(88,36,68,87,57,62,20,72, 3,46,33,67,46,55,12,32,63,93,53,69,),
( 4,42,16,73,38,25,39,11,24,94,72,18, 8,46,29,32,40,62,76,36,),
(20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74, 4,36,16,),
(20,73,35,29,78,31,90, 1,74,31,49,71,48,86,81,16,23,57, 5,54,),
( 1,70,54,71,83,51,54,69,16,92,33,48,61,43,52, 1,89,19,67,48,),
)
def seqs(nums, row, col):
if row + 4 <= len(nums): yield list(nums[i][col] for i in xrange(row, row+4))
if col + 4 <= len(nums[row]): yield list(nums[row][i] for i in xrange(col, col+4))
if row + 4 <= len(nums) and col + 4 <= len(nums[row]): yield list(nums[row+i][col+i] for i in xrange(0,4))
if row + 4 <= len(nums) and col >= 3: yield list(nums[row+i][col-i] for i in xrange(0,4))
def product(seq):
n = 1
for x in seq: n = n * x
return n
def list_seqs(nums):
for row in xrange(0, len(nums)):
for col in xrange(0, len(nums[row])):
for seq in seqs(nums, row, col):
yield seq
print max(product(seq) for seq in list_seqs(nums))
|
| 12 | 76576500 | 7593 | Show solution'''
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:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
'''
import prime
for i in xrange(1, 1000000000):
n = i * (i+1) / 2
if prime.num_factors(n) > 500:
print n
break
|
| 13 | 5537376230 | 141 | Show solution'''
Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690
'''
print str(sum((
37107287533902102798797998220837590246510135740250,
46376937677490009712648124896970078050417018260538,
74324986199524741059474233309513058123726617309629,
91942213363574161572522430563301811072406154908250,
23067588207539346171171980310421047513778063246676,
89261670696623633820136378418383684178734361726757,
28112879812849979408065481931592621691275889832738,
44274228917432520321923589422876796487670272189318,
47451445736001306439091167216856844588711603153276,
70386486105843025439939619828917593665686757934951,
62176457141856560629502157223196586755079324193331,
64906352462741904929101432445813822663347944758178,
92575867718337217661963751590579239728245598838407,
58203565325359399008402633568948830189458628227828,
80181199384826282014278194139940567587151170094390,
35398664372827112653829987240784473053190104293586,
86515506006295864861532075273371959191420517255829,
71693888707715466499115593487603532921714970056938,
54370070576826684624621495650076471787294438377604,
53282654108756828443191190634694037855217779295145,
36123272525000296071075082563815656710885258350721,
45876576172410976447339110607218265236877223636045,
17423706905851860660448207621209813287860733969412,
81142660418086830619328460811191061556940512689692,
51934325451728388641918047049293215058642563049483,
62467221648435076201727918039944693004732956340691,
15732444386908125794514089057706229429197107928209,
55037687525678773091862540744969844508330393682126,
18336384825330154686196124348767681297534375946515,
80386287592878490201521685554828717201219257766954,
78182833757993103614740356856449095527097864797581,
16726320100436897842553539920931837441497806860984,
48403098129077791799088218795327364475675590848030,
87086987551392711854517078544161852424320693150332,
59959406895756536782107074926966537676326235447210,
69793950679652694742597709739166693763042633987085,
41052684708299085211399427365734116182760315001271,
65378607361501080857009149939512557028198746004375,
35829035317434717326932123578154982629742552737307,
94953759765105305946966067683156574377167401875275,
88902802571733229619176668713819931811048770190271,
25267680276078003013678680992525463401061632866526,
36270218540497705585629946580636237993140746255962,
24074486908231174977792365466257246923322810917141,
91430288197103288597806669760892938638285025333403,
34413065578016127815921815005561868836468420090470,
23053081172816430487623791969842487255036638784583,
11487696932154902810424020138335124462181441773470,
63783299490636259666498587618221225225512486764533,
67720186971698544312419572409913959008952310058822,
95548255300263520781532296796249481641953868218774,
76085327132285723110424803456124867697064507995236,
37774242535411291684276865538926205024910326572967,
23701913275725675285653248258265463092207058596522,
29798860272258331913126375147341994889534765745501,
18495701454879288984856827726077713721403798879715,
38298203783031473527721580348144513491373226651381,
34829543829199918180278916522431027392251122869539,
40957953066405232632538044100059654939159879593635,
29746152185502371307642255121183693803580388584903,
41698116222072977186158236678424689157993532961922,
62467957194401269043877107275048102390895523597457,
23189706772547915061505504953922979530901129967519,
86188088225875314529584099251203829009407770775672,
11306739708304724483816533873502340845647058077308,
82959174767140363198008187129011875491310547126581,
97623331044818386269515456334926366572897563400500,
42846280183517070527831839425882145521227251250327,
55121603546981200581762165212827652751691296897789,
32238195734329339946437501907836945765883352399886,
75506164965184775180738168837861091527357929701337,
62177842752192623401942399639168044983993173312731,
32924185707147349566916674687634660915035914677504,
99518671430235219628894890102423325116913619626622,
73267460800591547471830798392868535206946944540724,
76841822524674417161514036427982273348055556214818,
97142617910342598647204516893989422179826088076852,
87783646182799346313767754307809363333018982642090,
10848802521674670883215120185883543223812876952786,
71329612474782464538636993009049310363619763878039,
62184073572399794223406235393808339651327408011116,
66627891981488087797941876876144230030984490851411,
60661826293682836764744779239180335110989069790714,
85786944089552990653640447425576083659976645795096,
66024396409905389607120198219976047599490197230297,
64913982680032973156037120041377903785566085089252,
16730939319872750275468906903707539413042652315011,
94809377245048795150954100921645863754710598436791,
78639167021187492431995700641917969777599028300699,
15368713711936614952811305876380278410754449733078,
40789923115535562561142322423255033685442488917353,
44889911501440648020369068063960672322193204149535,
41503128880339536053299340368006977710650566631954,
81234880673210146739058568557934581403627822703280,
82616570773948327592232845941706525094512325230608,
22918802058777319719839450180888072429661980811197,
77158542502016545090413245809786882778948721859617,
72107838435069186155435662884062257473692284509516,
20849603980134001723930671666823555245252804609722,
53503534226472524250874054075591789781264330331690,
)))[0:10]
|
| 14 | 837799 | 9092 | Show solution'''
The following iterative sequence is defined for the set of positive integers
n -> n/2 (n is even)
n -> 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
It can be seen that this sequence (starting at 13 and fiinishing at 1) contains 10 terms.
Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
'''
cache = { 1: 1 }
def chain(cache, n):
if not cache.get(n,0):
if n % 2: cache[n] = 1 + chain(cache, 3*n + 1)
else: cache[n] = 1 + chain(cache, n/2)
return cache[n]
m,n = 0,0
for i in xrange(1, 1000000):
c = chain(cache, i)
if c > m: m,n = c,i
print n
|
| 15 | 137846528820 | 141 | Show solution'''
Starting in the top left corner of a 2x2 grid, there are 6 routes (without backtracking) to the bottom right corner.
How many routes are there through a 20x20 grid?
'''
def fact(n):
f = 1
for x in xrange(1, n+1): f = f * x
return f
print fact(40) / fact(20) / fact(20)
|
| 16 | 1366 | 141 | Show solution'''
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26
What is the sum of the digits of the number 2^1000?
'''
def digits(n):
s = 0
while n > 0:
s = s + (n % 10)
n = n / 10
return s
print digits(pow(2,1000))
|
| 17 | 21124 | 203 | Show solution'''
How many letters would be needed to write all the numbers in words from 1 to 1000?
'''
words = [
( 1, 'one' , '' ),
( 2, 'two' , '' ),
( 3, 'three' , '' ),
( 4, 'four' , '' ),
( 5, 'five' , '' ),
( 6, 'six' , '' ),
( 7, 'seven' , '' ),
( 8, 'eight' , '' ),
( 9, 'nine' , '' ),
( 10, 'ten' , '' ),
( 11, 'eleven' , '' ),
( 12, 'twelve' , '' ),
( 13, 'thirteen' , '' ),
( 14, 'fourteen' , '' ),
( 15, 'fifteen' , '' ),
( 16, 'sixteen' , '' ),
( 17, 'seventeen', '' ),
( 18, 'eighteen' , '' ),
( 19, 'nineteen' , '' ),
( 20, 'twenty' , '' ),
( 30, 'thirty' , '' ),
( 40, 'forty' , '' ),
( 50, 'fifty' , '' ),
( 60, 'sixty' , '' ),
( 70, 'seventy' , '' ),
( 80, 'eighty' , '' ),
( 90, 'ninety' , '' ),
( 100, 'hundred' , 'and' ),
(1000, 'thousand' , 'and' ),
]
words.reverse()
def spell(n, words):
word = []
while n > 0:
for num in words:
if num[0] <= n:
div = n / num[0]
n = n % num[0]
if num[2]: word.append(' '.join(spell(div, words)))
word.append(num[1])
if num[2] and n: word.append(num[2])
break
return word
print sum(len(word) for n in xrange(1, 1001) for word in spell(n, words))
|
| 18 | 1074 | 437 | Show solution'''
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
7 5
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
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)
'''
triangle = (
(75, ),
(95, 64, ),
(17, 47, 82, ),
(18, 35, 87, 10, ),
(20, 4, 82, 47, 65, ),
(19, 1, 23, 75, 3, 34, ),
(88, 2, 77, 73, 7, 63, 67, ),
(99, 65, 4, 28, 6, 16, 70, 92, ),
(41, 41, 26, 56, 83, 40, 80, 70, 33, ),
(41, 48, 72, 33, 47, 32, 37, 16, 94, 29, ),
(53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14, ),
(70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57, ),
(91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48, ),
(63, 66, 4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31, ),
( 4, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 4, 23, ),
)
def path(triangle, num):
s = triangle[0][0]
col = 0
for row in xrange(1, len(triangle)):
if num % 2: col = col + 1
num = num / 2
s = s + triangle[row][col]
return s
print max(path(triangle, n) for n in xrange(0, 16384))
|
| 19 | 171 | 171 | Show solution'''
You are given the following information, but you may prefer to do some research for yourself.
1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
'''
import datetime
sundays = 0
for year in xrange(1901, 2001):
for month in xrange(1, 13):
d = datetime.date(year, month, 1)
if d.weekday() == 6:
sundays = sundays + 1
print sundays
|
| 20 | 648 | 140 | Show solution'''
Find the sum of digits in 100!
'''
def digits(n):
s = 0
while n > 0:
s = s + (n % 10)
n = n / 10
return s
n = 1
for i in xrange(1,100): n = n*i
print digits(n)
|
| 21 | 31626 | 8625 | Show solution'''
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
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.
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.
Evaluate the sum of all the amicable numbers under 10000.
'''
def divisors(n): return list(i for i in xrange(1, n/2+1) if n % i == 0)
pair = dict( ((n, sum(divisors(n))) for n in xrange(1, 10000)) )
print sum(n for n in xrange(1, 10000) if pair.get(pair[n], 0) == n and pair[n] != n)
|
| 22 | 871198282 | 203 | Show solution'''
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.
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.
What is the total of all the name scores in the file?
'''
def worth(name): return sum(ord(letter) - ord('A') + 1 for letter in name)
names = open('names.txt').read().replace('"', '').split(',')
names.sort()
print sum((i+1) * worth(names[i]) for i in xrange(0, len(names)))
|
| 23 | 4179871 | 15281 | Show solution'''
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.
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.
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.
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
'''
import prime
MAX = 28124
prime._refresh(MAX/2)
abundants = [n for n in xrange(1, MAX) if sum(prime.all_factors(n)) > n+n]
abundants_dict = dict.fromkeys(abundants, 1)
total = 0
for n in xrange(1, MAX):
sum_of_abundants = 0
for a in abundants:
if a > n: break
if abundants_dict.get(n - a):
sum_of_abundants = 1
break
if not sum_of_abundants:
total = total + n
print total
|
| 24 | 2783915460 | 157 | Show solution'''
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:
012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
'''
def fact(n):
f = 1
for x in xrange(1, n+1): f = f * x
return f
def permutation(orig_nums, n):
nums = list(orig_nums)
perm = []
while len(nums):
divider = fact(len(nums)-1)
pos = n / divider
n = n % divider
perm.append(nums[pos])
nums = nums[0:pos] + nums[pos+1:]
return perm
print ''.join(str(x) for x in permutation(range(0,10), 999999))
|
| 25 | 4782 | 140 | Show solution'''
What is the first term in the Fibonacci sequence to contain 1000 digits
'''
import math
phi = (1 + pow(5, 0.5)) / 2
c = math.log10(5) / 2
logphi = math.log10(phi)
n = 1
while True:
if n * logphi - c >= 999:
print n
break
n = n + 1
|
| 26 | 983 | 1421 | Show solution'''
Find the value of d < 1000 for which 1 / d contains the longest recurring cycle
'''
def cycle_length(n):
i = 1
if n % 2 == 0: return cycle_length(n / 2)
if n % 5 == 0: return cycle_length(n / 5)
while True:
if (pow(10, i) - 1) % n == 0: return i
else: i = i + 1
m = 0
n = 0
for d in xrange(1,1000):
c = cycle_length(d)
if c > m:
m = c
n = d
print n
|
| 27 | -59231 | 11108 | Show solution'''
Euler published the remarkable quadratic formula:
n^2 + n + 41
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.
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.
Considering quadratics of the form:
n^2 + an + b, where |a| <= 1000 and |b| <= 1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |4| = 4
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.
'''
import prime
max_pair = (0,0,0)
for a in xrange(-999, 1000):
for b in xrange(max(2, 1-a), 1000): # b >= 2, a + b + 1 >= 2
n, count = 0, 0
while True:
v = n*n + a*n + b
prime._refresh(v)
if prime.isprime(v): count = count + 1
else: break
n = n + 1
if count > max_pair[2]:
max_pair = (a,b,count)
print max_pair[0] * max_pair[1]
|
| 28 | 669171001 | 141 | Show solution'''
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
43 49
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
37 31
57
1
6*4
19*4
39*4
69*4
It can be verified that the sum of both diagonals is 101.
What is the sum of both diagonals in a 1001 by 1001 spiral formed in the same way?
'''
diagonal = 1
start = 1
for width in xrange(3, 1002, 2):
increment = width - 1
count = increment * 4
diagonal = diagonal + start * 4 + increment * 10
start = start + count
print diagonal
|
| 29 | 9183 | 233 | Show solution'''
How many distinct terms are in the sequence generated by ab for 2 <= a <= 100 and 2 <= b <= 100
'''
terms = {}
count = 0
for a in xrange(2,101):
for b in xrange(2,101):
c = pow(a,b)
if not terms.get(c, 0):
terms[c] = 1
count = count + 1
print count
|
| 30 | 443839 | 1703 | Show solution'''
Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
1634 = 1^4 + 6^4 + 3^4 + 4^4
8208 = 8^4 + 2^4 + 0^4 + 8^4
9474 = 9^4 + 4^4 + 7^4 + 4^4
As 1 = 1^4 is not a sum it is not included.
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
'''
def power_of_digits(n, p):
s = 0
while n > 0:
d = n % 10
n = n / 10
s = s + pow(d, p)
return s
print sum(n for n in xrange(2, 200000) if power_of_digits(n, 5) == n)
|
| 31 | 73682 | 230828 | Show solution'''
In England the currency is made up of pound, P, and pence, p, and there are eight coins in general circulation:
1p, 2p, 5p, 10p, 20p, 50p, P1 (100p) and P2 (200p).
It is possible to make P2 in the following way:
1 P1 + 1 50p + 2 20p + 1 5p + 1 2p + 3 1p
How many different ways can P2 be made using any number of coins?
'''
coins = (1, 2, 5, 10, 20, 50, 100, 200)
def balance(pattern): return sum(coins[x]*pattern[x] for x in xrange(0, len(pattern)))
def gen(pattern, coinnum, num):
coin = coins[coinnum]
for p in xrange(0, num/coin + 1):
newpat = pattern[:coinnum] + (p,)
bal = balance(newpat)
if bal > num: return
elif bal == num: yield newpat
elif coinnum < len(coins)-1:
for pat in gen(newpat, coinnum+1, num):
yield pat
print sum(1 for pat in gen((), 0, 200))
|
| 32 | 45228 | 43656 | Show solution'''
The product 7254 is unusual, as the identity, 39 x 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
'''
from combinatorics import permutations
def num(l):
s = 0
for n in l: s = s * 10 + n
return s
product = {}
for perm in permutations(range(1,10)):
for cross in range(1,4): # Number can't be more than 4 digits
for eq in range(cross+1, 6): # Result can't be less than 4 digits
a = num(perm[0:cross])
b = num(perm[cross:eq])
c = num(perm[eq:9])
if a * b == c: product[c] = 1
print sum(p for p in product)
|
| 33 | 100 | 171 | Show solution'''
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.
We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
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.
If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
'''
def fractions():
for numerator in map(str, xrange(10, 100)):
for denominator in map(str, xrange(int(numerator)+1, 100)):
if numerator == denominator: continue
if numerator[1] == denominator[1] and numerator[1] == '0': continue
if numerator[0] == denominator[0] and int(numerator) * int(denominator[1]) == int(denominator) * int(numerator[1]): yield(int(numerator), int(denominator))
if numerator[0] == denominator[1] and int(numerator) * int(denominator[0]) == int(denominator) * int(numerator[1]): yield(int(numerator), int(denominator))
if numerator[1] == denominator[1] and int(numerator) * int(denominator[0]) == int(denominator) * int(numerator[0]): yield(int(numerator), int(denominator))
if numerator[1] == denominator[0] and int(numerator) * int(denominator[1]) == int(denominator) * int(numerator[0]): yield(int(numerator), int(denominator))
def gcd(a,b): return b and gcd(b, a % b) or a
numerator = 1
denominator = 1
for frac in fractions():
numerator = numerator * frac[0]
denominator = denominator * frac[1]
g = gcd(numerator, denominator)
print denominator / g
|
| 34 | 40730 | 671 | Show solution'''
145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
Find the sum of all numbers which are equal to the sum of the factorial of their digits.
Note: as 1! = 1 and 2! = 2 are not sums they are not included.
'''
fact = (1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880)
def sum_of_digits_factorial(n):
s = 0
while n > 0:
d = n % 10
s = s + fact[d]
n = n / 10
return s
print sum(n for n in xrange(10, 100000) if n == sum_of_digits_factorial(n))
|
| 35 | 55 | 5406 | Show solution'''
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
'''
sieve = [True] * 1000000
sieve[0] = sieve[1] = False
def mark(sieve, x):
for i in xrange(x+x, len(sieve), x):
sieve[i] = False
for x in xrange(2, int(len(sieve) ** 0.5) + 1):
mark(sieve, x)
def circular(n):
digits = []
while n > 0:
digits.insert(0, str(n % 10))
n = n / 10
for d in xrange(1, len(digits)):
yield int(''.join(digits[d:] + digits[0:d]))
count = 0
for n, p in enumerate(sieve):
if p:
iscircularprime = 1
for m in circular(n):
if not sieve[m]:
iscircularprime = 0
break
if iscircularprime:
count = count + 1
print count
|
| 36 | 872187 | 753 | Show solution'''
The decimal number, 585 = 10010010012 (binary), is palindromic in both bases.
Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
(Please note that the palindromic number, in either base, may not include leading zeros.)
'''
# Faster solution by AbsoluteZero
print sum(n for n in xrange(1, 1000000)
if str(n)[::-1] == str(n) and bin(n)[2:] == bin(n)[-1:1:-1])
# My original solution
#
# def ispalindrome(n, base):
# digits = []
# reverse = []
# while n > 0:
# d = str(n % base)
# digits.append(d)
# reverse.insert(0, d)
# n = n / base
# return digits == reverse
#
# print sum(n for n in xrange(1, 1000000) if ispalindrome(n, 10) and ispalindrome(n, 2))
|
| 37 | 748317 | 234 | Show solution'''
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.
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
'''
import math, prime
digits = range(0, 10)
prime_digits = (2, 3, 5, 7)
def num(l):
s = 0
for n in l: s = s * 10 + n
return s
def is_left_truncatable(l):
is_truncatable = 1
for size in xrange(1, len(l)+1):
n = num(l[:size])
prime._refresh(int(math.sqrt(n)))
if not prime._isprime(n):
is_truncatable = 0
break
return is_truncatable
def is_right_truncatable(l):
is_truncatable = 1
for size in xrange(0, len(l)):
n = num(l[size:])
prime._refresh(int(math.sqrt(n)))
if not prime._isprime(n):
is_truncatable = 0
break
return is_truncatable
def gen(result, number):
if len(number) > 6: return
number = list(number)
number.append('')
for digit in digits:
number[-1] = digit
if is_left_truncatable(number):
if is_right_truncatable(number) and len(number) > 1:
result.append(num(number))
gen(result, number)
result = []
gen(result, [])
print sum(result)
|
| 38 | 932718654 | 359 | Show solution'''
Take the number 192 and multiply it by each of 1, 2, and 3:
192 x 1 = 192
192 x 2 = 384
192 x 3 = 576
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)
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).
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?
'''
def get_pandigital(n):
pandigital = ''
for x in xrange(1, 10):
pandigital += str(x * n)
if len(pandigital) >= 9: break
if len(pandigital) == 9 and sorted(dict.fromkeys(list(pandigital)).keys()) == list("123456789"): return pandigital
else: return ''
max = 0
for n in xrange(1, 10000):
p = get_pandigital(n)
if p and p > max: max = p
print max
|
| 39 | 840 | 22046 | Show solution'''
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.
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p < 1000, is the number of solutions maximised?
'''
maxp, maxsol = 0, 0
for p in xrange(12, 1001, 2):
solutions = 0
# a < b < c. So a is at most 1/3 of p. b is between a and (p-a)/2
for a in xrange(1, p/3):
a2 = a*a
for b in xrange(a, (p-a)/2):
c = p - a - b
if a2 + b*b == c*c: solutions = solutions + 1
if solutions > maxsol: maxp, maxsol = p, solutions
print maxp
|
| 40 | 210 | 157 | Show solution'''
An irrational decimal fraction is created by concatenating the positive integers:
0.123456789101112131415161718192021...
It can be seen that the 12th digit of the fractional part is 1.
If dn represents the nth digit of the fractional part, find the value of the following expression.
d1 x d10 x d100 x d1000 x d10000 x d100000 x d1000000
0 digit < 1
1 digit < + 9 * 1 10
2 digit < + 90 * 2 190
3 digit < + 900 * 3 2890
4 digit < + 9000 * 4
5 digit < + 90000 * 5
'''
def digit_at(n):
digits = 1
n = n - 1
while True:
numbers = 9 * pow(10, digits-1) * digits
if n > numbers: n = n - numbers
else: break
digits = digits + 1
num = n / digits + pow(10, digits-1)
return int(str(num)[n % digits])
print digit_at(1) * digit_at(10) * digit_at(100) * digit_at(1000) * digit_at(10000) * digit_at(100000) * digit_at(1000000)
|
| 41 | 7652413 | 155 | Show solution'''
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.
What is the largest n-digit pandigital prime that exists?
'''
import prime
from combinatorics import permutations
# Pan-digital primes are 4 or 7 digits. Others divisible by 3
prime._refresh(2766) # sqrt(7654321)
for perm in permutations(range(7, 0, -1)):
num = 0
for n in perm: num = num * 10 + n
if prime._isprime(num):
print num
break
|
| 42 | 162 | 187 | Show solution'''
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:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
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.
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?
'''
def worth(word): return sum(ord(letter) - ord('A') + 1 for letter in word)
words = open('words.txt').read().replace('"', '').split(',')
triangle_numbers = dict.fromkeys(list(n*(n+1)/2 for n in xrange(1, 100)), 1)
print sum(1 for word in words if worth(word) in triangle_numbers)
|
| 43 | 16695334890 | 10156 | Show solution'''
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.
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
d2 d3 d4 = 406 is divisible by 2
d3 d4 d5 = 063 is divisible by 3
d4 d5 d6 = 635 is divisible by 5
d5 d6 d7 = 357 is divisible by 7
d6 d7 d8 = 572 is divisible by 11
d7 d8 d9 = 728 is divisible by 13
d8 d9 d10= 289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
'''
from combinatorics import permutations
def num(l):
s = 0
for n in l: s = s * 10 + n
return s
def subdiv(l, n): return num(l) % n == 0
total = 0
for perm in permutations((0,1,2,3,4,6,7,8,9)):
perm.insert(5, 5) # d6 must be 5
if (subdiv(perm[7:10], 17) and
subdiv(perm[6:9], 13) and
subdiv(perm[5:8], 11) and
subdiv(perm[4:7], 7) and
subdiv(perm[3:6], 5) and
subdiv(perm[2:5], 3) and
subdiv(perm[1:4], 2)):
total += num(perm)
print total
|
| 44 | 5482660 | 4046 | Show solution'''
Pentagonal numbers are generated by the formula, P_n=n(3n-1)/2. The first ten pentagonal numbers are:
1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
It can be seen that P_4 + P_7 = 22 + 70 = 92 = P_8. However, their difference, 70 - 22 = 48, is not pentagonal.
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?
'''
MAX = 2000
pent = [ n * (3*n - 1) / 2 for n in xrange(1, 2*MAX) ]
pdic = dict.fromkeys(pent)
def main2():
for j in xrange(0, MAX):
for k in xrange(j+1, 2*MAX-1):
p_j = pent[j]
p_k = pent[k]
p_sum = p_j + p_k
p_diff = p_k - p_j
if pdic.has_key(p_sum) and pdic.has_key(p_diff):
return p_diff
print main2()
|
| 45 | 1533776805 | 812 | Show solution'''
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle T_n=n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal P_n=n(3n-1)/2 1, 5, 12, 22, 35, ...
Hexagonal H_n=n(2n-1) 1, 6, 15, 28, 45, ...
It can be verified that T_285 = P_165 = H_143 = 40755.
Find the next triangle number that is also pentagonal and hexagonal.
'''
MAX = 100000
triangle = [ n * ( n + 1) / 2 for n in xrange(0, MAX) ]
pentagon = [ n * (3*n - 1) / 2 for n in xrange(0, MAX) ]
hexagon = [ n * (2*n - 1) for n in xrange(0, MAX) ]
pentagon_dict = dict.fromkeys(pentagon, 1)
hexagon_dict = dict.fromkeys(hexagon, 1)
for t in xrange(286, MAX):
v = triangle[t]
if pentagon_dict.has_key(v) and hexagon_dict.has_key(v):
print v
break |
| 46 | 5777 | 437 | Show solution'''
It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
9 = 7 + 2 x 1^2
15 = 7 + 2 x 2^2
21 = 3 + 2 x 3^2
25 = 7 + 2 x 3^2
27 = 19 + 2 x 2^2
33 = 31 + 2 x 1^2
It turns out that the conjecture was false.
What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
'''
import prime
MAX = 10000
squares = dict.fromkeys((x*x for x in xrange(1, MAX)), 1)
prime._refresh(MAX)
for x in xrange(35, MAX, 2):
if not prime.isprime(x):
is_goldbach = 0
for p in prime.prime_list[1:]:
if p >= x: break
if squares.has_key((x - p)/2):
is_goldbach = 1
break
if not is_goldbach:
print x
break
|
| 47 | 134043 | 90046 | Show solution'''
Find the first four consecutive integers to have four distinct prime factors
'''
import prime
def distinct_factors(n): return len(dict.fromkeys(prime.factors(n)).keys())
factors = [0, 1, distinct_factors(2), distinct_factors(3)]
while True:
if factors[-4::] == [4,4,4,4]: break
else: factors.append(distinct_factors(len(factors)))
print len(factors)-4
|
| 48 | 9110846700 | 344 | Show solution'''
Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
'''
s = 0
mod = pow(10, 10)
for x in xrange(1, 1001):
s = s + pow(x, x)
print s % mod
|
| 49 | 296962999629 | 515 | Show solution'''
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.
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.
What 12-digit number do you form by concatenating the three terms in this sequence?
'''
import prime
from combinatorics import permutations
prime._refresh(10000)
for num in xrange(1000, 10000):
if str(num).find('0') >= 0: continue
if prime.isprime(num):
prime_permutations = { num: 1 }
for x in permutations(list(str(num))):
next_num = int(''.join(x))
if prime.isprime(next_num):
prime_permutations[next_num] = 1
primes = sorted(prime_permutations.keys())
for a in xrange(0, len(primes)):
if primes[a] == 1487: continue
for b in xrange(a+1, len(primes)):
c = (primes[a] + primes[b]) / 2
if prime_permutations.has_key(c):
print str(primes[a]) + str(c) + str(primes[b])
exit()
|
| 50 | 997651 | 10625 | Show solution'''
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
'''
import prime
MAX = 5000
prime.prime(MAX)
def check_length(n, below):
maxprime = 0
for x in xrange(0, below):
total = sum(prime.prime_list[x:x+n])
if total > below: break
if prime.isprime(total): maxprime = total
return maxprime
for n in xrange(1000, 0, -1):
maxprime = check_length(n, 1000000)
if maxprime:
print maxprime
break
|
| 51 | 121313 | 34968 | Show solution'''
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.
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.
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.
'''
import prime
from combinatorics import uniqueCombinations
cache = {}
def prime_family_length(n, digits):
if cache.has_key((n, digits)): return cache[n, digits]
num, nums, count = list(str(n)), [], 0
if len(dict.fromkeys(num[d] for d in digits).keys()) > 1:
return cache.setdefault((n, digits), 0) # The digits must have the same number
for d in range(0 in digits and 1 or 0, 10): # Ensure 0 is not the first digit
for x in digits: num[x] = str(d)
n = int(''.join(num))
if prime.isprime(n): count += 1
nums.append(n)
for n in nums: cache[n, digits] = count
return count
prime._refresh(100000)
n, max, max_count, combos = 10, 0, 0, {}
while max_count < 8:
p = prime.prime(n)
digits = range(0, len(str(p)))
for size in xrange(1, len(digits)):
patterns = combos.setdefault((len(digits), size),
tuple(tuple(sorted(p)) for p in uniqueCombinations(digits, size)))
for pat in patterns:
count = prime_family_length(p, pat)
if count > max_count: max, max_count = p, count
n += 1
print p
|
| 52 | 142857 | 640 | Show solution'''
It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
'''
def multiples_have_same_digits(n):
digit_keys = dict.fromkeys(list(str(n)))
for x in xrange(2, 4):
for d in list(str(x * n)):
if not digit_keys.has_key(d): return False
return True
n = 0
while True:
n = n + 9 # n must be a multiple of 9 for this to happen
if multiples_have_same_digits(n):
print n
break
|
| 53 | 4075 | 280 | Show solution'''
There are exactly ten ways of selecting three from five, 12345:
123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
In combinatorics, we use the notation, 5C3 = 10.
In general,
nCr = n! / r!(nr)! where r <= n, n! = n x (n-1)...x 3 x 2 x 1, and 0! = 1.
It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
How many, not necessarily distinct, values of nCr, for 1 <= n <= 100, are greater than one-million?
'''
fact_c = { 0: 1, 1: 1 }
def fact(n): return fact_c.has_key(n) and fact_c[n] or fact_c.setdefault(n, n * fact(n-1))
count = 0
for n in xrange(1, 101):
for r in xrange(0, n):
ncr = fact(n) / fact(r) / fact(n-r)
if ncr > 1000000: count += 1
print count
|
| 54 | 376 | 860 | Show solution'''
In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:
High Card: Highest value card.
One Pair: Two cards of the same value.
Two Pairs: Two different pairs.
Three of a Kind: Three cards of the same value.
Straight: All cards are consecutive values.
Flush: All cards of the same suit.
Full House: Three of a kind and a pair.
Four of a Kind: Four cards of the same value.
Straight Flush: All cards are consecutive values of same suit.
Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.
The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
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.
Consider the following five hands dealt to two players:
Hand Player 1 Player 2 Winner
1 5H 5C 6S 7S KD 2C 3S 8S 8D TD Player 2
Pair of Fives Pair of Eights
2 5D 8C 9S JS AC 2C 5C 7D 8S QH Player 1
Highest card Ace Highest card Queen
3 2D 9C AS AH AC 3D 6D 7D TD QD Player 2
Three Aces Flush with Diamonds
4 4D 6S 9H QH QC 3D 6D 7H QD QS Player 1
Pair of Queens Pair of Queens
Highest card Nine Highest card Seven
5 2H 2D 4C 4D 4S 3C 3D 3S 9S 9D Player 1
Full House Full House
With Three Fours with Three Threes
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.
How many hands does Player 1 win?
'''
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 }
all_kinds = tuple(reversed(sorted(value.values())))
all_suits = list('DCSH')
def make_hand(cards):
hand = {}
for card in cards:
hand.setdefault(value[card[0]], {})[card[1]] = 1
hand.setdefault(card[1], {})[value[card[0]]] = 1
return hand
def get(hash, arr): return ((i, hash.get(i, {})) for i in arr)
def has(hash, arr): return not sum(1 for i in arr if i not in hash)
def rank(hand):
# Royal flush
for suit, kinds in get(hand, all_suits):
if has(kinds, tuple('TJQKA')):
return (9,)
# Straight flush
for suit, kinds in get(hand, all_suits):
kinds = sorted(kind for kind in kinds.keys())
if len(kinds) == 5 and kinds[4] - kinds[0] == 4:
return (8, kinds[0])
# Four of a kind
for kind, suits in get(hand, all_kinds):
if len(suits.keys()) == 4:
return (7, kind)
# Full house
for kind, suits in get(hand, all_kinds):
if len(suits.keys()) == 3:
for kind2, suits2 in get(hand, all_kinds):
if len(suits2.keys()) == 2:
return (6, kind, kind2)
# Flush
for suit, kinds in get(hand, all_suits):
if len(kinds.keys()) == 5:
return (5,)
# Straight
kinds = sorted(kind for kind in all_kinds if hand.has_key(kind))
if len(kinds) == 5 and kinds[4] - kinds[0] == 4:
return (4, kinds[0])
# Three of a kind
for kind, suits in get(hand, all_kinds):
if len(suits.keys()) == 3:
return (3, kind)
# Two pairs
for kind, suits in get(hand, all_kinds):
if len(suits.keys()) == 2:
for kind2, suits2 in get(hand, all_kinds):
if kind != kind2 and len(suits2.keys()) == 2:
return (2, kind, kind2)
# One pair
for kind, suits in get(hand, all_kinds):
if len(suits.keys()) == 2:
return (1, kind)
for kind in all_kinds:
if kind in hand:
return (0, kind)
return 0
count = 0
for hand in open('poker.txt'):
hands = hand.split(' ')
p1, p2 = make_hand(hands[0:5]), make_hand(hands[5:10])
v1, v2 = rank(p1), rank(p2)
if v1 > v2: count += 1
print count
|
| 55 | 249 | 546 | Show solution'''
If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
Not all numbers produce palindromes so quickly. For example,
349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337
That is, 349 took three iterations to arrive at a palindrome.
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).
Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.
How many Lychrel numbers are there below ten-thousand?
NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.
'''
def is_lychrel(n):
n = str(n)
for count in xrange(0, 50):
n = str(int(n) + int(n[::-1]))
if n == n[::-1]: return False
return True
print sum(1 for n in xrange(0, 10000) if is_lychrel(n))
|
| 56 | 972 | 1922 | Show solution'''
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.
Considering natural numbers of the form, a^b, where a, b < 100, what is the maximum digital sum?
'''
max = 0
for a in xrange(0, 100):
for b in xrange(0, 100):
ds = sum(int(digit) for digit in str(a**b))
if ds > max: max = ds
print max |
| 57 | 153 | 328 | Show solution'''
It is possible to show that the square root of two can be expressed as an infinite continued fraction.
sqrt(2) = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
By expanding this for the first four iterations, we get:
1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
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.
In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
'''
num, den, count = 3, 2, 0
for iter in xrange(0, 1000):
num, den = num + den + den, num + den
if len(str(num)) > len(str(den)):
count += 1
print count
|
| 58 | 26241 | 6952 | Show solution'''
Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 5 4 3 12 29
40 19 6 1 2 11 28
41 20 7 8 9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49
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%.
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%?
'''
import prime
prime._refresh(50000)
width, diagonal, base, primes = 1, 1, 1, 0
while True:
width = width + 2
increment = width - 1
for i in xrange(0, 4):
diagonal = diagonal + increment
if i < 3 and prime._isprime(diagonal): primes += 1
base = base + 4
if primes * 10 < base:
print width
break
|
| 59 | 107359 | 9203 | Show solution'''
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.
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.
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.
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.
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.
'''
from combinatorics import selections
code = tuple(int(c) for c in open('cipher1.txt').read().split(','))
def decrypt(code, password):
l = len(password)
return tuple(c ^ password[i % l] for i, c in enumerate(code))
def text(code): return ''.join(chr(c) for c in code)
n = 0
for password in selections(tuple((ord(c) for c in list('abcdefghijklmnopqrstuvwxyz'))), 3):
c = decrypt(code, password)
t = text(c)
if t.find(' the ') > 0:
print sum(c)
break |