Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it United States

Problem:

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?

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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