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Problem 1168

AIME late

Combinatorics Difficulty 5.2 Prove it Harvard-MIT Mathematics Tournament · United States

The numbers $112$ , $121$ , $123$ , $153$ , $243$ , $313$ , and $322$ are among the rows, columns, and diagonals of a $3 \times 3$ square grid of digits (rows and diagonals read left-to-right, and columns read top-to-bottom). What 3-digit number completes the list?

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:

1	1	2
5	2	4
3	1	3

The center digit is the middle digit of 4 numbers (hence at least 3 members of the above list), so it must be

2

. The top-left digit begins at least 2 members of the above list, so it must be

1

3

. If it is

3

, then after placing

313

we see that we need three more numbers starting with

3

, impossible; hence, it is

1

. So

243

and

313

must (in some order) be the last row and the last column, and now it is easy to complete the grid as shown; the answer is

524

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.