Nthe sequence $121, 1221, 12221, \dots ,$ the $n$th number consists of $n$ copies of the digit $2$, surrounded by two $1$s. how many of the first $100$ terms in the sequence are divisible by $3$?

The th number in the sequence can be expressed as

Extracting the th term from the sum gives

and . 3 divides both 12 and 21, so and 21 contribute no remainder.

This leaves us with

Recall that a decimal integer is divisible by 3 if its digits add to a multiple of 3. The digits in are copies of 2 and one 0, so the digital sum is .

This means that roughly 1/3 of the first numbers in this sequence are divisible by 3; among the first 100 terms, they occur for , of which there are 33.

  33

Step-by-step explanation:

The number is divisible by 3 if the sum of digits is divisible by 3. 1221 is the first number with a sum of digits divisible by 3. Adding 3 more 2s to the sequence will result in another number divisible by 3.

So, every 3rd term from n=2 to n=98 will be divisible by 3, for a total of 33 terms out of the first 100 in the sequence.

95.03 is the answer. hope this !