Project Euler With Python

Tackling problems from project Euler using Python

Project Euler With Python

Project Euler is a website set up with a wide selection of math problems. While some, if not all of them, could be figured out manually over time you could be looking at taking weeks or months to find an answer.

The idea is that you use any form of coumputer programming you are happy with to create a programme that returns the answer. The challenge is knowing how to structure and write the programme to complete the task at hand. When done correctly a computer can return the answer instantly or in seconds.

The puzzles can all be found at https://projecteuler.net and if you create an account you can select a puzzle and check if you answer is correct.

My Solutions

My full workings on this project, including answers and runtimes, are available to view here on my Github page as Jupyter notebooks including more detailed notes on each problem: https://github.com/seanmerricks/Python_Project_Euler


Problem 1

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.
import time
start = time.time()
answer = 0

for x in range(1, 1000):
    if x % 3 == 0 or x % 5 == 0:
        answer = answer + x
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 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:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
import time
start = time.time()
answer = 0
x = 0
y = 1
z = 0

while z < 4000000:
    z = x + y
    if z % 2 == 0:
        answer = answer + z
    x = y
    y = z
    
elapsed = float("{0:.2f}".format((time.time() - start)))       
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 3

The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
import time
start = time.time()
answer = 0
number = 600851475143
x = 2

while x < number ** 0.5:
    while number % x == 0:
        number = number / x
    x += 1

answer = int(number)

elapsed = float("{0:.2f}".format((time.time() - start)))       
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 4

A palindromic number reads the same both ways.
The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
import time
start = time.time()
answer = 0
working = 0
highest = 0
x = 999
y = 999

while answer == 0 and x > 1 and y > 1:
    while y > 1:
        working = str(x * y)
        if working == working[::-1] and int(working) > highest:
            highest = int(working)
            y -= 1
        else:
            y -= 1
    x -= 1
    y = 999
    
answer = highest

elapsed = float("{0:.2f}".format((time.time() - start)))       
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 5

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 positive number that is evenly divisible by all of the numbers from 1 to 20?
import time
start = time.time()
answer = 0
x = 1
y = 20

while answer == 0:
    if x % y == 0:
        if y == 1:
            answer = x
        else:
            y -= 1
    else:
        x += 1
        y = 20

elapsed = float("{0:.2f}".format((time.time() - start)))       
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 6

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 = 55 2 = 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.
import time
start = time.time()
answer = 0
sumsq = 0
sumxsq = 0
 
for x in range(1,101):
    sumsq += x ** 2
    sumxsq += x

sumxsq = sumxsq ** 2
answer = sumxsq - sumsq
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 7

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 10 001st prime number?
import time
import math
start = time.time()
answer = 0
number = 10001
working = 0
x = 2


def prime_check(num):
    """
    Input: Any number.
    Output: A true or false response to state if the number is prime.
    Notes: The function only checks up to the square root of the test number and skips all even numbers after 2 to save time.
    """
    if num % 2 == 0 and num > 2: 
        return False
    for i in range(3, int(math.sqrt(num)) + 1, 2):
        if num % i == 0:
            return False
    return True

while working < number:
    if prime_check(x) == True:
        working += 1
        answer = x
    x += 1  
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 8

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
import time
start = time.time()
answer = 0
working = 1
adjust = 13
number = "73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698747158523863050715693290963295227443043557668966489504452445231617318564030987111217223831136222989342338030813533627661428280644448664523874930358907296290491560440772390713810158593079608667017242712188399879790879227492190169972088809377665727333001053367881220235421809751254540594752243525849077116705560136048395864467063244157221553975369781797784617406495514929086256932197846862248283972241375657056057490261407972968652414535100474821663704844031998900088952434506585412275886668811642717147992444292823063465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"

for y in range(1, (len(number) - adjust)):
    for x in range(1, adjust + 1):
        working = working * int(number[(y + x):(y + x) + 1])
    if working > answer:
        answer = working
    working = 1
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 9

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.
import time
start = time.time()
answer = 0

for x in range(1,500):
    for y in range(1,500):
        z = ((x ** 2) + (y ** 2)) ** 0.5
        if z - int(z) == 0:
            a = x + y + z
            if a == 1000:
                answer = x * y * z
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",int(answer), "calculated in",elapsed,"seconds.")

Problem 10

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.

This one could do with some level of rework before I would be happy with it as it took a little over 8 mins to run all around and I like to keep them under the minute mark as per the website guidelines. I mean it still works but there is obviously a cleaner way to do this.

import time
import math
start = time.time()
answer = 0

def prime_check(num):
    """
    Input: Any number.
    Output: A true or false response to state if the number is prime.
    Notes: The function only checks up to the square root of the test number and skips all even numbers after 2 to save time.
    """
    if num % 2 == 0 and num > 2: 
        return False
    for i in range(3, int(math.sqrt(num)) + 1, 2):
        if num % i == 0:
            return False
    return True

for x in range(2,2000000):
    if prime_check(x) == True:
        answer += x
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 11

n the 20×20 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 × 63 × 78 × 14 = 1788696.
What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
import time
start = time.time()
answer = 0

my_list =  [[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]]

for y in range(0,20):
    for x in range(0,17):
        z = my_list[y][x] * my_list[y][x + 1] * my_list[y][x + 2] * my_list[y][x + 3]
        if z > answer:
            answer = z
    
for y in range(0,20):
    for x in range(0,17):
        z = my_list[x][y] * my_list[x + 1][y] * my_list[x + 2][y] * my_list[x + 3][y]
        if z > answer:
            answer = z

for y in range(0,17):
    for x in range(0,17):
        z = my_list[y][x] * my_list[y + 1][x + 1] * my_list[y + 2][x + 2] * my_list[y + 3][x + 3]
        if z > answer:
            answer = z

for y in range(0,17):
    for x in range(3,20):
        z = my_list[y][x] * my_list[y + 1][x - 1] * my_list[y + 2][x - 2] * my_list[y + 3][x - 3]
        if z > answer:
            answer = z
        
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 12

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?

No doubt this one is a brute force effort and takes far to long to run repeatedly but it does give the correct soultion. Thinking again I could have it account for the square root of each triangle number and stop attempting to divide each number for factors when it crosses that point which should save time when working with such high numbers.

Still working on it :)

Problem 13

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
import time
start = time.time()
answer = 0
working = 0

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]

for num in numbers:
    answer += num
    working += 1
    
answer = str(answer)[0:10]
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 14

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 finishing 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.
import time
start = time.time()
answer = 0
working = 0

def collatz(num):
    """
    Input: Any number.
    Output: The response from the collatz sequence based on the input number.
    """
    if num % 2 == 0:
        return int((num / 2))
    elif num % 2 == 1:
        return int((3 * num + 1))

for x in range(1,1000000):
    steps = 1
    number = x
    while number > 1:
        number = collatz(number)
        steps += 1
    if steps > working:
        working = steps
        answer = x
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 15

Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.

How many such routes are there through a 20×20 grid?
Still working on it :)

Problem 16

215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 21000?
import time
start = time.time()
answer = 0

x = [int(num) for num in str(2 ** 1000)]

for num in x:
    answer += num
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 17

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
import time
from num2words import num2words
start = time.time()
answer = 0
my_list = []
my_string = ""

for x in range(1, 1001):
    my_string += num2words(x)

my_string = my_string.replace(" ", "")
my_string = my_string.replace("-", "")
print(my_string)
answer = len(my_string)

elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 19

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 time
start = time.time()
answer = 0
total_days = 0
days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

for x in range(1901, 2001):
    if x % 4 == 0:
        days[1] = 29
    else:
        days[1] = 28
    
    for y in range(0,12):
        if (total_days + 1) % 7 == 0:
            answer += 1
        total_days += days[y]
        total_days -= 1
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 20

n! means n × (n − 1) × ... × 3 × 2 × 1

For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!
import time
start = time.time()
answer = 0

def n_exclamation(number):
    """
    Input: Any number.
    Output: A value of the n! function by multiplying the given number by every number between itself and 1.
    """
    x = number
    working = 1
    while x > 0:
        working = working * x
        x -= 1
    return working

x = [int(num) for num in str(n_exclamation(100))]

for num in x:
    answer += num
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 22

Using names.txt *note , 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 × 53 = 49714.

What is the total of all the name scores in the file?

*note (available on the project Euler site, for the sake of this problem a copy is stored within the same directory as the Jupyter notebook)
import time
start = time.time()
answer = 0

with open('p022_names.txt', 'r') as my_file:
    names = my_file.read().split(',')

names = [s.strip('"') for s in names]
names.sort()

def name_value(name, iteration, temp_answer):
    for letter in name:
        temp_answer += ord(letter) - 64
    temp_answer = temp_answer * iteration
    return temp_answer

iteration = 1
answer = 0

for name in names:
    temp_answer = name_value(name, iteration, 0)
    iteration += 1
    answer += temp_answer
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 24

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?
import time
import itertools
start = time.time()
answer = 0

my_list = [0,1,2,3,4,5,6,7,8,9]

my_perms = set(itertools.permutations(my_list))
my_perms = sorted(my_perms)

answer = str(my_perms[999999])
answer = answer.replace(", ","")
answer = answer.replace("(","")
answer = answer.replace(")","")

elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 29

Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:

22 = 4, 23 = 8, 24 = 16, 25 = 32
32 = 9, 33 = 27, 34 = 81, 35 = 243
42 = 16, 43 = 64, 44 = 256, 45 = 1024
52 = 25, 53 = 125, 54 = 625, 55 = 3125

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
import time
start = time.time()
answer = 0

my_list = []

for x in range(2, 101):
    for y in range(2, 101):
        temp_answer = x ** y
        my_list.append(temp_answer)

my_set = set(my_list)
answer = len(my_set)
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 36

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.)
import time
start = time.time()
answer = 0

def is_palindrome(number):
    if str(number)[::-1] == str(number):
        return True
    else:
        return False

for x in range(1, 1000001):
    if is_palindrome(x) == True:
        temp_binary = "{0:b}".format(x)
        if is_palindrome(temp_binary) == True:
            answer += x
    
elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 39

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?
import time
import math
start = time.time()
answer = 0
length = 0

my_list = []

def tri_perim(p):
    temp_list = []
    for a in range(1, p - 1):
        for b in range(1, p - 1):
            c = math.sqrt((a ** 2) + (b ** 2))
            if c.is_integer() == True:
                if (a + b + c) == p:
                    temp_list.append([a,b,c])
    for item in temp_list:
        item.sort()
    if len(temp_list) > answer:
        return [p, len(set(map(tuple, temp_list)))]
    else:
        return [0,0]
    
for x in range(3, 1001):
    tri_test = tri_perim(x)
    if tri_test[1] > length:
        answer = x
        length = tri_test[1]

elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 48

The series, 11 + 22 + 33 + ... + 1010 = 10405071317.

Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.
import time
start = time.time()
answer = 0

for x in range(1, 1001):
    answer += x ** x

answer = str(answer)
y = len(answer) - 10
answer = answer[y::]

elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 55

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.
import time
start = time.time()
answer = 0
check = [False]

def lychrel_check(number, iterations):
    temp_number = number + int(str(number)[::-1])
    if str(temp_number)[::-1] == str(temp_number):
        return [True, temp_number, iterations]
    else:
        iterations += 1
        return [False, temp_number, iterations]

for x in range(1, 10001):
    iterations = 0
    check = lychrel_check(x, iterations)
    while ((check[0] != True) and (iterations < 49)):        
        check = lychrel_check(check[1], iterations)
        number = check[1]
        iterations += 1
        if ((check[0] == False) and (iterations == 49)):
            answer += 1

elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 56

A googol (10100) is a massive number: one followed by one-hundred zeros; 100100 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, ab, where a, b < 100, what is the maximum digital sum?
import time
start = time.time()
answer = 0

def digit_sum(a, b):
    temp_number = a ** b
    temp_list = sum([int(x) for x in str(temp_number)])
    return temp_list
    
#digit_sum(5, 100)

for x in range(1, 101):
    for y in range(1, 101):
        temp_answer = digit_sum(x, y)
        if temp_answer > answer:
            answer = temp_answer

elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 74

The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:

1! + 4! + 5! = 1 + 24 + 120 = 145

Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:

169 → 363601 → 1454 → 169
871 → 45361 → 871
872 → 45362 → 872

It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,

69 → 363600 → 1454 → 169 → 363601 (→ 1454)
78 → 45360 → 871 → 45361 (→ 871)
540 → 145 (→ 145)

Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.

How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
import time
import math
start = time.time()
answer = 0
dupe = False
my_list = []

for x in range(1, 1000001):
    dupe = False
    num = x
    my_list.append(x)
    while (dupe == False) and (len(my_list) < 61):
        temp_num = sum([math.factorial(int(char)) for char in str(num)])
        if temp_num not in my_list:
            my_list.append(temp_num)
        else:
            dupe = True
        num = temp_num
    if len(my_list) == 60:
        answer += 1
    my_list = []

elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")

Problem 92

A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.

For example,

44 → 32 → 13 → 10 → 1 → 1

85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89

Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.

How many starting numbers below ten million will arrive at 89?
import time
start = time.time()
answer = 0

def number_chain(number):
    temp = 0
    for x in str(number):
        temp += int(x) ** 2
    return temp

for x in range(1, 10000001):
    chain_temp = number_chain(x)
    while chain_temp > 1:
        if chain_temp == 1:
            break
        elif chain_temp == 89:
            answer += 1
            break
        else:
            chain_temp = number_chain(chain_temp)

elapsed = float("{0:.2f}".format((time.time() - start)))
print("Answer:",answer, "calculated in",elapsed,"seconds.")