Maths Olympiad Prep

Track / Stage 5 / 210 of 400 #810 of 1964

Problem 810

AIME late
Combinatorics Difficulty 5.5 Find the answer

1. In the cells of a 10×1010 \times 10 table, the numbers 1,2,3,,1001,2,3, \ldots, 100 are arranged such that the sum of the numbers in any 2×22 \times 2 square does not exceed SS. Find the smallest possible value of SS.

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Answer: 202.

Solution. Divide the 10×1010 \times 10 table into 25 squares of 2×22 \times 2. Since the sum of the numbers in the entire table is

1+2++100=1001012=5050 1+2+\cdots+100=\frac{100 \cdot 101}{2}=5050

the arithmetic mean of the sums of the numbers in these 25 squares is 202. Therefore, in at least one square, the sum of the numbers is not less than 202, that is, S202S \geqslant 202. An example of an arrangement where the value S=202S=202 is achieved is shown in the figure.

100999897969594939291
12345678910
90898887868584838281
11121314151617181920
80797877767574737271
21222324252627282930
70696867666564636261
31323334353637383940
60595857565554535251
41424344454647484950

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