Track / Stage 5 / 360 of 400 #960 of 2000

Problem 960

AIME late

Geometry Difficulty 5.9 Prove it

# 8.5. (7 points)

A large square was cut out from graph paper along the grid lines. From it, a smaller square was also cut out along the grid lines. After this, exactly 79 cells remained from the large square. Was it necessary for the cut-out square to contain one of the corner cells of the large square?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Answer: Yes, definitely.

Solution: Let the large square contain $N^{2}$ cells, and the cut-out square - $M^{2}$ cells, then $N^{2}-M^{2}=79$ , that is, $(N-M)(N+M)=79$ .

Since 79 is a prime number, then $\left\{\begin{array}{c}N+M=79 \\ N-M=1\end{array}\right.$ . The solution to this system is $N=40, M=39$ . Therefore, the cut-out square was adjacent to two sides of the larger one, that is, it contained one of its corner cells.

## Comment.

A fully justified solution - 7 points.

A generally correct reasoning is provided, with minor gaps or inaccuracies - 5 points.

Only the correct answer is provided - 1 point.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.