The four adjacent digits in the $1000$-digit number that have the greatest product are $9 \cdot 9 \cdot 8 \cdot 9 = 5832$.
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?
The problem asks for the maximum product of 13 consecutive digits inside a large 1000-digit number. A simple brute-force method would multiply every possible group of 13 consecutive digits, but we can make it more efficient by noting that any window containing a zero will yield a product of zero. Therefore, we can split the number into zero-free segments and compute products only within those segments.
x = """73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450"""
x = x.replace("\n", "")
max_prod = 0
for segment in x.split('0'):
if len(segment) < 13:
continue
digits = list(map(int, segment))
for i in range(len(digits) - 13 + 1):
prod = 1
for d in digits[i:i+13]:
prod *= d
max_prod = max(max_prod, prod)
print(max_prod)
Explained line by line:
max_prod to keep track
of the maximum product found so far, starting at zero.digits.prod to 1 and multiplying each digit in the group.max
_prod if the current product is greater than the previously recorded maximum.Running the above code leads to an answer of $23514624000$