Sum of Product of Elements of Subsets

Level 2

Consider S, the set of all positive integers from 1 to N. For each non-empty subset of S, calculate the product of the elements of the subset. Let the sum of all such products be f(N). For example, since the non-empty subsets of {1,2,3} are {1}, {2}, {3}, {1,2}, {2,3}, {1,3} and {1,2,3}, f ( 3 ) = 1 + 2 + 3 + 1 × 2 + 2 × 3 + 1 × 3 + 1 × 2 × 3 = 23 f(3)=1+2+3+1 \times 2+2 \times 3+1 \times 3+1 \times 2 \times 3=23 What is the minimum value of N such that f(N)+1 is a multiple of 1000?


The answer is 14.

Joel Tan by Joel Tan , 21, Singapore

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1 solution

Erika Wang
Mar 25, 2018

f ( n ) + 1 = ( 1 + 1 ) ( 1 + 2 ) ( 1 + 3 ) ⋯ ( 1 + n ) = ( n + 1 ) ! f(n)+1=(1+1)(1+2)(1+3)\cdots(1+n)=(n+1)! Then ( n + 1 ) ! (n+1)! must have three fives in its prime factorization, and testing multiples of 5 5 starting with 5 , 5, we see that n + 1 = 1 ⟹ n = 14 . n+1=1\implies n=\boxed{14}.

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