Let A = {27, 51, 62, 73, 84, 11, 54}. How many subsets of s

Question

Let A = {27, 51, 62, 73, 84, 11, 54}.

How many subsets of size 5 of A have an average that is an integer?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 10

sanyam900 · Answer

sol? i didnt understand why im wrong here. also if theres an easier way.

cabecudo · Answer

If the average is an integer, then the sum of the subset must be divisible by 5.

Creating a subset of 5 elements from a set of 7 is the same as removing 2 elements from the original set.

Because we are tracking if the subset is divisible by 5, we can work with a set which consists only of the remainders of the elements when divided by 5:
{27, 51, 62, 73, 84, 11, 54} -> {2, 1, 2, 3, 4, 1, 4}.

The sum of this new set is 2+1+2+3+4+1+4 = 17, whose remainder when divided by 5 is 2.

This means that the 2 elements we are removing from this set must have a sum whose remainder is also 2. If we do that, the sum of the subset will be a number divisible by 5, thus the average of the subset will be an integer.

In our set of remainders, the only pairs that sum to 2 or 7 are (1,1) and (3,4). There is only 1 way to choose the pair {1,1}, and since there is one 3 and two 4s, there are 2 ways to choose a {3,4} pair.

Thus, there are 1 + 2 = 3 subsets of size 5 of A that have an average that is an integer.

Option (B)

I don`t usually write solutions so I hope it`s clear enough

PGTLrowanhand · Answer

While you could add all the numbers up to see what the average is, it will be–as noted–much easier just to look at how much each group deviates from a multiple of 5.

We need to make sure that  the sum of these differences from 5  are also a multiple of 5 if we want an integer answer.

Here’s the process:

(I tend to do this all in positives so I don’t have to worry about minus signs.)

So if we go through the numbers {27, 51, 62, 73, 84, 11, 54} that gives us (all as pluses): {2, 1, 2, 3, 4, 1, 4}

Now we have to consider the groups that are possible.

Given this set of numbers, we’ll either have a sum of 10 or a sum of 15 given that all seven numbers here only add to 17 total, and there aren’t many ways to get either 10 or 15.

Sum of 10: {2,1,2,3,2} or {2,1,2,4,1}  
 Sum of 15: {2,2,3,4,4}

Those are your three options. The answer is (B).

Let A = {27, 51, 62, 73, 84, 11, 54}. How many subsets of s

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Let A = {27, 51, 62, 73, 84, 11, 54}. How many subsets of s

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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