Question

Let $S=\{1,2,3,......,40\}$ and let A be a subset of S such that no two elements in A have their sum divisible by 5. What is the maximum number of elements possible in A?
A. 10
B. 13
C. 17
D. 20

Answer 1

Hint: When we are going to check the sum of two elements in A is divisible by 5 then it is too complicated. Instead of this we can check whether the unit digit of sum of two elements is 0 or 5. If the unit digit of the sum of two elements is 0 or 5 then according to the divisibility test the sum is divisible by 5.

Complete step-by-step answer:

We have been given that Set $S=\{1,2,3,.....,40\}$. We have to find out those elements for which the sum of the pair of elements is not divisible by 5. So, we have to check whether the sum of pairs of elements has unit digit 0 or 5 or not.
Now, we can take all elements of Set S which have unit digit 1 and 2 because if the unit digit is 1 and 2 then the unit digit of the sum of the pair of elements is not equal to 0 or 5. So, elements having unit digit 1 and 2 are; 1, 2, 11, 12, 21, 22, 31, 32.
Here we can’t take those elements of Set S which have unit digit 3 and 4 because when the unit digit is 3 then when it is added to elements having the unit digit 2, we get unit digit of sum is 5 and similarly unit digit 4 added with unit digit 1 we get unit digit 5. So, for 3 and 4 we get the sum as unit digit 5 which is divisible by 5.
Now, we can take elements having the unit digit 6 and 7 because if the unit digit is 1, 2, 6, and 7 then the sum of pairs of elements don’t have unit digit 0 or 5. So, elements having unit digits 6 and 7 are; 6, 7, 16, 17, 26, 27, 36, 37.
Here we can’t take those elements of Set S which has unit digit 8 and 9 because when the unit digit is 8 then when it is added to elements having the unit digit 2 or 7, we get unit digit of sum is 0 or 5 and similarly unit digit 9 added with unit digit 1 or 6 and get unit digit 0 or 5. So, for 8 and 9 we get the sum as unit digit 0 or 5 which is divisible by 5.
Now, we can take any one element having unit digit 0 or 5 because they add up with any elements having unit digit 1, 2, 6, 7 we won’t get unit digit 0 or 5.
So, now we are collecting all elements of Set A.
$A=\left\{ \begin{align}
  & 1,11,21,31 \\
 & 2,21,22,32 \\
 & 6,16,26,36 \\
 & 7,17,27,37 \\
\end{align}\right\}$+$\text{ any one element of }\left\{ \begin{align}
  & 5,15,25,35 \\
 & 10,20,30,40 \\
\end{align} \right\}$
So, the total maximum number of elements in set A is 17.
Hence, the correct option for the given question is option (c).

Note: The possibility of mistake is we can’t take one element of Set $\left\{ \begin{align}
  & 5,15,25,35 \\
 & 10,20,30,40 \\
\end{align} \right\}$ because all these elements are divisible by 5. But here we are talking about the sum of any two elements which is not getting 0 or 5.