Bunuel wrote:

Set M consists of all positive integers that are multiples of 4. Set N consists of all positive integers less than 100 that have a units digit of 8. How many integers do sets M and N have in common?

(A) Three

(B) Four

(C) Five

(D) Six

(E) Seven

Essentially the question asks how may elements in set M are multiples of 4

So Set M={8,18,28...........,88,98}.

as there are only 11 elements in set M, we can actually count the numbers that are divisible by 4

they are 8,28,48,68 and 88 i.e. 5 elements. Hence Option

C----------------------------------

Alternatively, first number of set M that is divisible by 4 is 8, next number will be 28 and last number that is divisible by 4 is 88. This is an AP series with

\(a=8\), \(d=28-8=20\) and \(T_n=88\),

Hence, \(T_n=a+(n-1)*d=>88=8+(n-1)*20=>n=5\)