Olympiad Maths Prep

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Combinatorics Difficulty 5.2 AIME, harder Prove it Ukraine

Oleksii placed positive integers in the cells of the chessboard of size 8×88 \times 8. For each pair of adjacent cells, Fedir wrote down the product of the numbers in them and added all the obtained numbers. Oleksii wrote down the sum of the numbers in each pair of adjacent cells and multiplied all the obtained numbers. It turned out that both numbers have the same last digit, 11. Prove that at least one of the boys made a mistake in the calculation. For example, for the square 3×33 \times 3 and the given arrangement of numbers (fig. 11), Fedir would write the following numbers: 22, 66, 88, 2424, 1515, 3535, 22, 66, 88, 2020, 1818, 4242 and their sum ends with digit 66, and Oleksii would write the following numbers: 33, 55, 66, 1010, 88, 1212, 33, 55, 66, 99, 99, 1313 and their product ends with digit 00.

(Oleksii Masalitin, Fedir Yudin)

| 1 | 2 | 3 |
|---|---|---|
| 2 | 4 | 6 |
| 3 | 5 | 7 |

Fig. 11

Solution

Suppose that this could happen. Since Oleksii's product is odd, all the factors are odd, and therefore the sum of the numbers in any neighboring cells is odd. But then for any neighboring cells, the parity of the numbers in those cells is different, and therefore the product of those numbers is even, and therefore the sum is even. Contradiction.

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