(Q). Consider a set s = {1, 2, 3, ………., 200}, two elements p and q are selected from this set s such that 7ᵖ + 7ᑫ is divisible by 5. In how many ways this selection can be done?
Solution:-
For a number to be divisible by 5, its last digit i.e. unit digit should be either 0 or 5 and unit digit of 7ⁿ can be 1, 3, 7 or 9
type of n Unit digit
7⁴ˣ 1
7⁴ˣ⁺¹ 7
7⁴ˣ⁺² 9
7⁴ˣ⁺³ 3
Hence from any combination of above unit digit taken two at a time last 5 can not be acheived but we can acheive last digit as 0 in the following way:-
p q
4x 4x + 2
4x + 2 4x
4x + 3 4x + 1
4x + 1 4x + 3
Here in set s = {1, 2, 3, 4, 5, ……………, 200}
Number of numbers of 4x type = 50
Number of number of 4x + 1 type = 50
Number of number of 4x + 2 type = 50
Number of number of 4x + 3 type = 50
∴ required number of ways of selection
= ⁵⁰C₁×⁵⁰C₁ + ⁵⁰C₁×⁵⁰C₁ + ⁵⁰C₁×⁵⁰C₁ + ⁵⁰C₁×⁵⁰C₁
= 4×⁵⁰C₁×⁵⁰C₁
= 4×50×50
= 10000 ways Answer