In the sequence 121, 1221, 12221,...., the nth number consists of n copies of the digit 2, surrounded by two 1s. How many of the first 100 terms in the sequence are divisible by 3?
The terms that will be divisible by 3 are the
2nd, 5th, 8th, 11th etc.
So.....
2 + 3(n - 1) ≤ 100
2 + 3n - 1 ≤ 100
-1 + 3n ≤ 100
3n ≤ 101
n ≤ 101/3
n = 33 terms
