Project Euler: Problem 8

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

(see the number below)

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

Use NumPy for this computation



In [1]:

    
import numpy as np



In [2]:

    
d1000 = """
73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586\
17866458359124566529476545682848912883142607690042\
24219022671055626321111109370544217506941658960408\
07198403850962455444362981230987879927244284909188\
84580156166097919133875499200524063689912560717606\
05886116467109405077541002256983155200055935729725\
71636269561882670428252483600823257530420752963450
"""

Turn d1000 into a list I can work with



In [3]:

    
b = []
for digit in d1000.strip():
    b.append(int(digit))
b









    Out[3]:





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

create a function that finds the max product of 13 adjacent digits



In [19]:

    
c = np.array(b)
def product(n):
    themax = 0
    numbers = []
    for i in range(1000):
        d = n[i:i+13]     #this is a list of the 13 adjacent digits
        p = np.cumprod(d)[len(np.cumprod(d)) - 1]      #this is the product of that list
        if p > themax:       #p and d will continue to be replaced by the largest that the for loop finds
            themax = p
            numbers = d
    return themax, numbers



In [20]:

    
print(product(c))









    



(23514624000, array([5, 5, 7, 6, 6, 8, 9, 6, 6, 4, 8, 9, 5]))

Success! Worked on this solution with Ryan Werth during pair programming in class.



In [ ]:

    
assert True # leave this for grading