Q.

The number of elements in the set {n∈ℕ:10≤n≤100 and 3n-3 is a multiple of 7} is __________.


Ans.

(15)

3n-3=7k,  k∈Z⇒3n=7k+3,  k∈Z                   ...(i)

Now, 3≡3 mod 7

33≡-1 mod 7

36≡1 mod 7

37≡3 mod 7

313≡3 mod 7

∴  n=1,7,13,… will satisfy equation (i).

This forms an A.P. with common difference 6.

Now,   n∈[10,100]

∴  n=13,19,…

Now,  13+(n-1)·6≤100⇒13+6n-6≤100

⇒6n≤100-7⇒6n≤93⇒n≤15.5⇒n=15

So, there are 15 such numbers.