The number of elements in the set {n∈ℕ:10≤n≤100 and 3n-3 is a multiple of 7} is __________ . [2023]
(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.