Can you properly distribute Candies?

Ω = f ( 2015 , 1 ) + f ( 2015 , 4 ) + f ( 2015 , 7 ) + + f ( 2015 , 4030 ) \large{\Omega = f(2015,\ 1) + f(2015,\ 4) + f(2015,\ 7) + \ldots + f(2015,\ 4030)}

Let f ( n , k ) f(n,\ k) be the number of ways of distributing k k indistinguishable candies to n n distinguishable children so that each child receives at most two candies.

For example, if n = 3 n=3 , then f ( 3 , 7 ) = 0 ; f ( 3 , 6 ) = 1 f(3,\ 7) = 0; f(3,\ 6)=1 and f ( 3 , 4 ) = 6 f(3,\ 4) = 6 . If the value of Ω \Omega can be represented as A B \large{A^B} for positive integers A , B A,B , where A A isn't a perfect n t h n^{th} power of an integer with n > 1 n>1 , submit the value of A + B A+B as your answer.


The answer is 2017.

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1 solution

Ivan Koswara
Aug 5, 2015

Distribute the candies to the first 2014 2014 children freely; there are 3 3 choices for each child (zero, one, or two candies). There exists exactly one choice for the last child that will make the sum of candies congruent to 1 1 modulo 3 3 . All these configurations will count to Ω \Omega , and no other. This gives Ω = 3 2014 \Omega = \boxed{3^{2014}} , so A = 3 , B = 2014 , A + B = 2017 A = 3, B = 2014, A+B = 2017 .

Moderator note:

Great observation with setting up the bijection!

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