Olympiad Maths Prep

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

Problem 260

AMC 10/12, early questions
Number theory Difficulty 4.0 Find the answer

Let SS be a subset of {1,2,3,,30}\{1,2,3,\dots,30\} with the property that no pair of distinct elements in SS has a sum divisible by 55. What is the largest possible size of SS?
(A) 10(B) 13(C) 15(D) 16(E) 18\textbf{(A)}\ 10\qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18

Official solution

Of the integers from 11 to 3030, there are six each of 0,1,2,3,4 (mod 5)0,1,2,3,4\ (\text{mod}\ 5). We can create several rules to follow for the elements in subset SS. No element can be 1 (mod 5)1\ (\text{mod}\ 5) if there is an element that is 4 (mod 5)4\ (\text{mod}\ 5). No element can be 2 (mod 5)2\ (\text{mod}\ 5) if there is an element that is 3 (mod 5)3\ (\text{mod}\ 5). Thus we can pick 6 elements from either 1 (mod 5)1\ (\text{mod}\ 5) or 4 (mod 5)4\ (\text{mod}\ 5) and 6 elements from either 2 (mod 5)2\ (\text{mod}\ 5) or 3 (mod 5)3\ (\text{mod}\ 5) for a total of 6+6=126+6=12 elements. Considering 0 (mod 5)0\ (\text{mod}\ 5), there can be one element that is so because it will only be divisible by 55 if paired with another element that is 0 (mod 5)0\ (\text{mod}\ 5). The final answer is (B) 13\boxed{\textbf{(B)}\ 13}.

== Solution 2 == ( Similar to solution 1 but more detailed)
Since we are taking the sum of distinct numbers mod(5), the only values that can be the remainder is 00,11,22,33,44. We dismiss the case of remainder 00 because this would make the sum a multiple of 5, which is not allowed. In addition, since we can't have 0 or 5 mod(5), we make some restraints. There are four different cases for the "allowed" mods in the sequence, or the remainders that the sum of distinct elements are allowed to be in the sequence. They are:
1) 1mod(5), and 3mod(5)

2) 1mod(5), and 2mod(5)

3) 2mod(5), and 4mod(5),

4) 2mod(5), and 1mod(5)
Now, we can calculate the number of elements in each of these cases. For the first case, we rewrite 1mod(5) in the form 5k+1. Because 5k+1 < 30, which is the largest element in the sequence possible, we obtain k=5. In addition, for 3mod(5), we have 3x+5<30, x=5. This gives us 5+5=10 solutions. The reason why we can add these elements is because all of these elements, no matter which two you add, can never be a multiple of 5. The only possible remainders for the sum are 1+1, 1+3, or 3+3.
In a similar manner, we calculate the element using the same logic. We find that 2mod(5) has 5 elements, and 1mod(5) has 8 elements, meaning that there are a total of 13 elements, which is the answer.
~CharmaineMa07292010

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