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"What is Project Euler? Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.

The motivation for starting Project Euler, and its continuation, is to provide a platform for the inquiring mind to delve into unfamiliar areas and learn new concepts in a fun and recreational context."


I have written a walkthrough providing solutions in Python. For each problem I have written a series of hints for one particular approach to solving the problem. The final hint for each problem is my code for the solution. The Project Euler website states that each problem should be solved with code that runs in under a minute.
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Problem 1 - Multiples of 3 and 5

​Question - 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.
  • We need to loop through all the numbers from 1 to 1000

  • If a number is divisible by 3 or 5 then we should keep a running total

  • total = 0
    for sec in range(1000):
       if sec % 3 == 0:
          total = total + sec
       elif sec % 5 == 0:
          total = total + sec
    
    print total
     

  • The answer is 233168

  • Show Hint


    ​Problem 2 - Even Fibonacci numbers

    Questions - 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,...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.
    ​
  • The first step is to generate values of the Fibonacci sequence

  • Then check if the value is even, if it is add it to a running total

  • total = 2
    i=0
    a=1
    b=2
    while i == 0:
       c=a+b
       if c>4000000:
          i=1
       elif c%2==0:
          total = total + c
       a=b
       b=c
    print total
     

  • The answer is 4613732

  • Show Hint


    ​Problem 3 - Largest prime factor

    Question - The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?
  • A number is prime iff it is only evenly divisible by itself and 1

  • We need to increment through the possible factors of 600851475143 and check whether they are prime

  • def isprime(test):
       ind = 0
       sec = 2
          while sec < test:
             if test % sec ==0:
                ind = 1
          sec = sec + 1
             if ind == 0:
                return 1
             else:
                return 0
    
    largest = 1
    sec = 2
    current = 600851475143 
    while sec < current :
    if current % sec ==0:
          if isprime(sec)==1:
             largest = sec
             current = current /sec
             sec = 1
       sec = sec +1
    
    print largest
    
    

  • The answer is 6857

  • Show Hint


    Problem 4 - Largest palindrome product

    Question - A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009=91×99.9009=91×99.
    Find the largest palindrome made from the product of two 3-digit numbers.
  • There is no built in function to check if a number is a palindrome, so we need to write our own.

  • If you convert an integer into a string then it gives much more flexibility in picking out specific digits.

  • def ispalindrome(test):
        ind = 1
        test = str(test)
        length = len(test)
        sec = 0
        while sec < length:
            if test[sec] != test[length - sec - 1]:
                ind = 0
            sec = sec + 1
        return ind
    
    largest = 1
    for sec in range(1000):
        for sec1 in range(1000):
            if ispalindrome(sec * sec1) == 1:
                if sec * sec1 > largest:
                    largest = sec * sec1
    
    print largest
    
    

  • The answer is 906609

  • Show Hint


    Problem 5 - Smallest multiple

    Question - 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?
  • A number is prime iff it is only evenly divisible by itself and 1

  • We need to increment through the possible factors of 600851475143 and check whether they are prime

  • import time
    
    def evenlydivisible(test):
        ind = 1
        sec = 3
        while sec < 20:
            if test % sec != 0:
                ind = 0
                sec = 21
            sec = sec + 1
        return ind
    
    start = time.time()
    ind1 = 0
    sec1 = 32
    while ind1 == 0:
        if evenlydivisible(sec1) == 1:
            ind1 = 1
        sec1 = sec1 + 16
        #print sec1
    
    
    print sec1 -1
    duration = time.time() - start
    print duration
    

  • The answer is 6857

  • Show Hint


    Problem 6 - Sum square difference

    Question - The sum of the squares of the first ten natural numbers is,
    12+22+...+102=38512+22+...+102=385
    The square of the sum of the first ten natural numbers is,
    (1+2+...+10)2=552=3025(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=2640.3025−385=2640.
    Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
    ​
  • We need to loop through the numbers 1 to 100

  • As we are looping through we should keep two running totals - one adding up the squares of the number we are currently on, one adding up the number we are currently on.

  • import time
    
    start = time.time()
    sumofsquares = 0
    squareofsums = 0
    for sec in range(101):
        sumofsquares = sumofsquares + sec**2
        squareofsums = squareofsums + sec
    
    print squareofsums**2 - sumofsquares
    print time.time() - start
    
    

  • The answer is 6857

  • Show Hint


    Problem 7 - 10001st prime

    Question - 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?
  • There is no way of analytically deriving the next prime number, so we are going to have to work our way through the first 10001 prime numbers

  • In order to write an algorithm that runs in a reasonable length of time we will need to optimize our test of whether a number is prime or not.

  • To check if an integer is prime we only need to check that it does not have any factors up to the square root of the number we are checking. We can use this fact to speed up our algorithm

  • We also know that 2 is the only even prime, therefore when checking for primes we only need to check odd numbers

  • import time
    import math
    
    def isprime(test):
        ind = 0
        sec = 2
        while sec <= math.sqrt(test):
            if test % sec == 0:
                ind = 1
                sec = test
            sec = sec + 1
        if ind == 0:
            return 1
        else:
            return 0
    
    sec = 3
    numberofprimes = 1
    start = time.time()
    while numberofprimes < 10001:
        if isprime(sec) == 1:
            numberofprimes = numberofprimes + 1
        sec = sec + 2    
    
    print sec - 2
    print time.time() - start
    
    Show Hint


    Problem 8 - Largest product in a series

    Question - The four adjacent digits in the 1000-digit number that have the greatest product are 9×9×8×9=5832.9×9×8×9=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?
  • There is no way of analytically deriving the next prime number, so we are going to have to work our way through the first 10001 prime numbers

  • In order to write an algorithm that runs in a reasonable length of time we will need to optimize our test of whether a number is prime or not.

  • To check if an integer is prime we only need to check that it does not have any factors up to the square root of the number we are checking. We can use this fact to speed up our algorithm

  • We also know that 2 is the only even prime, therefore when checking for primes we only need to check odd numbers

  • import time
    
    number8 = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'
    
    start = time.time()
    largest = 1
    
    for sec in range(987):
        product = 1
        for sec1 in range(13):
            product = product * int(number8[sec+sec1])
        if product > largest:
            largest = product
    
    print largest
    print time.time() - start
    
    Show Hint


    Problem 9 - Special Pythagorean triplet

    ​Question - 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.
  • We can supply the integers a and b, then calculate c based on the fact that it is a Pythagorean Triplet.

  • import time
    
    def ispy(a,b,c):
        if a**2+b**2 == c**2:
            return 1
        else:
            return 0
    
    start = time.time()
    
    a = 1
    b = 1
    ind = 1
    while ind == 1:
        if b == 1001:
            b=1
            a=a+1
        c = 1000 - b - a
        if ispy(a,b,c) == 1:
            ind = 0
        b = b + 1
            
    print a *(b - 1)* c
    print time.time() - start
    
    Show Hint


    ​Problem 10 - Summation of primes 

    ​Question - 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 is quite straightforward, we just need to loop through the first 2 million integers, and check for each one if it is prime. We can even reuse our primality test from earlier problems.

  • import time
    import math
    
    def isprime(test):
        ind = 0
        if test % 3 == 0:
            ind = 1
        elif test % 5 == 0:
            ind = 1
        else:
            sec = 7
            while sec <= math.sqrt(test):
                if test % sec == 0:
                    ind = 1
                    sec = test
                sec = sec + 4
                if test % sec == 0:
                    ind = 1
                    sec = test
                sec = sec + 2
        if ind == 0:
            return 1
        else:
            return 0
    
    sec = 5
    sumofprimes = 10
    start = time.time()
    while sec < 2000001:
        if isprime(sec) == 1:
            sumofprimes = sumofprimes + sec
        sec = sec + 2 
    
    print sumofprimes
    print time.time() - start
    
    Show Hint


    Problem 11 - Largest product in a grid

    Question - In 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?
  • This one is quite straightforward, we just need to loop through the first 2 million integers, and check for each one if it is prime. We can even reuse our primality test from earlier problems.

  • import time
    
    def returnnumber(row,col):
        return a[row*60 + col*3:row*60 + col*3+2]
    
    start = time.time()
    
    a = "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"
    
    largest = 1
    
    for sec in range(20):
        for sec1 in range(17):
            new = 1
            for sec2 in range(4):
                new = new * int(returnnumber(sec,sec1 + sec2))
                if new > largest:
                    largest = new
    
    for sec in range(17):
        for sec1 in range(20):
            new = 1
            for sec2 in range(4):
                new = new * int(returnnumber(sec + sec2,sec1))
                if new > largest:
                    largest = new
    
    for sec in range(17):
        for sec1 in range(17):
            new = 1
            for sec2 in range(4):
                new = new * int(returnnumber(sec + sec2,sec1 + sec2))
                if new > largest:
                    largest = new
     
    for sec in range(17):
        for sec1 in range(17):
            new = 1
            for sec2 in range(4):
                new = new * int(returnnumber(sec + 3 - sec2,sec1 +sec2))
                if new > largest:
                    largest = new
                    
    print largest        
    print time.time() - start
    
    
    Show Hint


    Problem 12 - Highly divisible triangular number

    Question - 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: 1,2,4,7,14,28We 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?
  • This one is another fairly straightforward problem, we just need to generate the triangular numbrs sequentially, and count the factors.

  • import time
    import math
    
    def ntriangle(n):
        return n*(n+1)/2
    
    def maxfactor(test,div):
        sec = 0
        while div**sec < test:
            if test % div**(sec+1) == 0:
                sec = sec + 1
            else:
                test = 1
        return sec
    
    def numfactors(sec):
        nfactors = 1
        sec1 = 2
        while sec1 <= sec:
            if sec % sec1 == 0:
                mfactors = maxfactor(sec,sec1)
                sec = sec / (sec1**mfactors)
                nfactors = nfactors * (mfactors + 1)
            sec1 = sec1 + 1
        return nfactors
    
    start = time.time()
    
    nfactors = 1
    sec = 5
    while nfactors < 500:
        nfactors = numfactors(ntriangle(sec))
        sec = sec + 1
        #print sec, nfactors
    
    print ntriangle(sec - 1)
    print numfactors(ntriangle(sec - 1))
    
    print time.time() - start
    
    Show Hint


    Problem 13 - Large sum

    Question - 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
    import math
    
    start = time.time()
    
    f = open("problem 13 numbers.txt" ,"r+")
    m=[]
    sum = 0
    
    for sec in range(100):
        m.append(f.read(50))
        f.read(1)
    
    f.close()
    
    for sec in range(100):
        sum = sum + int(m[sec])
                
    print (sum - sum % (10**42))/(10**42)
    
    print time.time() - start
    Show Hint


    Problem 14 - Longest Collatz sequence

    Question - 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
    import math
    
    start = time.time()
    
    def lencollatz(sec):
        length = 1
        while sec != 1:
            if sec % 2 == 0:
                sec = sec / 2
            else:
                sec = 3 * sec + 1
            length = length + 1
        return length
    
    sec = 1
    longest = 1
    longestnum = 1
    while sec < 1000000:
        new = lencollatz(sec)
        if new > longestnum:
            longestnum = new
            longest = sec
        sec = sec + 1
    
    print longest, longestnum
        
    print time.time() - start
    Show Hint


    Problem 15 - Lattice paths

    Picture
    Question - 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?


    Problem 16 - Power digit sum

    Question - 215=32768215=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
    import math
    
    start = time.time()
    
    def sumdigit(sec):
        sec = str(sec)
        length = 0
        for sec1 in range(len(sec)):
            length = length + int(sec[sec1])
        return length
    
    print sumdigit(2**1000)
        
    print time.time() - start
    Show Hint


    Problem 17 - Number letter counts

    Question - 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?
    import time
    import math
    
    start = time.time()
    
    def sumdigit(sec):
        sec = str(sec)
        length = 0
        for sec1 in range(len(sec)):
            length = length + int(sec[sec1])
        return length
    
    print sumdigit(2**1000)
        
    print time.time() - start
    Show Hint
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