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Problem 1309

AIME late

Algebra Difficulty 5.4 Prove it HMMT November · United States

Consider a $2 \times 2$ grid of squares. David writes a positive integer in each of the squares. Next to each row, he writes the product of the numbers in the row, and next to each column, he writes the product of the numbers in each column. If the sum of the eight numbers he writes down is $2015$ , what is the minimum possible sum of the four numbers he writes in the grid?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Let the four numbers be $a, b, c, d$ , so that the other four numbers are $ab, ad, bc, bd$ . The sum of these eight numbers is $a + b + c + d + ab + ad + bc + bd = (a + c) + (b + d) + (a + c)(b + d) = 2015$ , and so $(a + c + 1)(b + d + 1) = 2016$ . Since we seek to minimize $a + b + c + d$ , we need to find the two factors of $2016$ that are closest to each other, which is easily calculated to be $42 \cdot 48 = 2016$ ; this makes $a + b + c + d = 88$ .

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