, 2013

Number theory Difficulty 5.9 AIME, harder Prove it Japan

Suppose two $20 \times 13$ rectangular grids consisting of $260$ small squares are given. We insert into each square box of the two grids, numbers $1, 2, \ldots, 260$ in the following way:
* For the first grid, we start inserting numbers $1, 2, \ldots, 13$ into the boxes on the top row from left to right. Continue to insert numbers $14, 15, \ldots, 26$ into the boxes on the second row from left to right. Keep on going until you finish inserting numbers $248, 249, \ldots, 260$ from left to right into the boxes on the bottom row.
* For the second grid, we start inserting numbers $1, 2, \ldots, 20$ into the boxes on the right-most column from top to bottom. Continue to insert numbers $21, 22, \ldots, 40$ into the boxes on the second column from the right from top to bottom. Keep on doing until you finish inserting numbers $241, 242, \ldots, 260$ into the boxes on the left most column from top to bottom.

List all the positive integers which get inserted into the boxes located in the same position in the two grids in the two ways of distributing numbers described above.

Solutions — 2

Solution 1

$87$ , $174$

The number inserted into the box located on the $i$ -th row from the top and $j$ -th column from the left is given by $13(i - 1) + j$ for the case of the first grid, and by $20(13 - j) + i$ for the case of the second grid. If the same number goes into the boxes located at the same position in the two grids, we must have $13(i - 1) + j = 20(13 - j) + i$ . Simplifying this identity, we get $12i + 21j = 273$ . Since both $21j$ and $273$ are multiples of $7$ , we must have $12i$ to be a multiple of $7$ , which implies that $i$ must be a multiple of $7$ . From the fact that $1 \le i \le 20$ , it follows that we get $i = 7, 14$ . The corresponding values for $j$ are $9$ , $5$ , respectively, and we see by computing $13(i - 1) + j = 20(13 - j) + i$ for these values of $i, j$ that $87$ and $174$ give the desired answer.

Solution 2

The number inserted into the box located on the $i$ -th row from the top and $j$ -th column from the left is given by $13(i-1)+j$ for the case of the first grid, and by $20(13-j)+i$ for the case of the second grid. If the same number goes into the boxes located at the same position in the two grids, we must have $13(i-1)+j = 20(13-j)+i$ . Simplifying this identity, we get $12i + 21j = 273$ . Since both $21j$ and $273$ are multiples of $7$ , we must have $12i$ to be a multiple of $7$ , which implies that $i$ must be a multiple of $7$ . From the fact that $1 \le i \le 20$ , it follows that we get $i=7, 14$ . The corresponding values for $j$ are $9, 5$ , respectively, and we see by computing $13(i-1)+j = 20(13-j)+i$ for these values of $i, j$ that $87$ and $174$ give the desired answer.

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