2008 iTest Problems/Problem 19

Problem

Let $A$ be the set of positive integers that are the product of two consecutive integers. Let $B$ be the set of positive integers that are the product of three consecutive integers. Find the sum of the two smallest elements of $A \cap B$ .

Solution

The first few numbers in set $B$ are $6,24,60,120,210,336 \cdots$ . To see if these numbers equal the product of two consecutive numbers, note that $n(n+1) = n^2+n$ and check perfect squares just below these values. After guessing and checking, the first values that equal the product of two consecutive numbers are $2 \cdot 3 = 6$ and $14 \cdot 15 = 210$ , and the sum of these two equal $\boxed{216}$ .

2008 iTest (Problems)
Preceded by: Problem 18	Followed by: Problem 20
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2008 iTest Problems/Problem 19

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