Olympiad Maths Prep

Track / Stage 3 / 144 of 260 #144 of 2000

Problem 144

AMC 10/12, early questions
Number theory Difficulty 3.6 Find the answer

Let SS be a subset of {1,2,3,...,50}\{1,2,3,...,50\} such that no pair of distinct elements in SS has a sum divisible by 77. What is the maximum number of elements in SS?
(A) 6(B) 7(C) 14(D) 22(E) 23\text{(A) } 6\quad \text{(B) } 7\quad \text{(C) } 14\quad \text{(D) } 22\quad \text{(E) } 23

Official solution

The fact that x0mod77xx \equiv 0 \mod 7 \Rightarrow 7 \mid x is assumed as common knowledge in this answer.
First, note that there are 88 possible numbers that are equivalent to 1mod71 \mod 7, and there are 77 possible numbers equivalent to each of 22-6mod76 \mod 7.
Second, note that there can be no pairs of numbers aa and bb such that aba \equiv -b mod 77, because then a+b7a+b | 7. These pairs are (0,0)(0,0), (1,6)(1,6), (2,5)(2,5), and (3,4)(3,4). Because (0,0)(0,0) is a pair, there can always be 11 number equivalent to 0mod70 \mod 7, and no more.
To maximize the amount of numbers in S, we will use 11 number equivalent to 0mod70 \mod 7, 88 numbers equivalent to 11, and 1414 numbers equivalent to 22-55. This is obvious if you think for a moment. Therefore the answer is 1+8+14=231+8+14=23 numbers.
E\fbox{E}

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