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HMMT February

Combinatorics Difficulty 5.0 AIME, harder Find the answer

Find the number of solutions in positive integers (k;a1,a2,…,ak;b1,b2,…,bk)(k ; a_{1}, a_{2}, \ldots, a_{k} ; b_{1}, b_{2}, \ldots, b_{k}) to the equation a1(b1)+a2(b1+b2)+⋯+ak(b1+b2+⋯+bk)=7a_{1}(b_{1})+a_{2}(b_{1}+b_{2})+\cdots+a_{k}(b_{1}+b_{2}+\cdots+b_{k})=7

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let k,a1,…,ak,b1,…,bkk, a_{1}, \ldots, a_{k}, b_{1}, \ldots, b_{k} be a solution. Then b1,b1+b2,…,b1+⋯+bkb_{1}, b_{1}+b_{2}, \ldots, b_{1}+\cdots+b_{k} is just some increasing sequence of positive integers. Considering the aia_{i} as multiplicities, the aia_{i} 's and bib_{i} 's uniquely determine a partition of 7. Likewise, we can determine aia_{i} 's and bib_{i} 's from any partition of 7, so the number of solutions is p(7)=15p(7)=15.

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