Question

n the 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 $3$?

Answer 1

The
`n`th number in the sequence can be expressed as

`a_n=10^(n+1)+\displaystyle2\sum_(i=2)^n10^i+21`

Extracting the
`n`th term from the sum gives

`a_n=10^(n+1)+2\cdot10^n+\displaystyle2\sum_(i=2)^(n-1)10^i+21`

and
`10^(n+1)+2\cdot10^n=10^n(10+2)=12\cdot10^n`. 3 divides both 12 and 21, so
`12\cdot10^n` and 21 contribute no remainder.

This leaves us with

`a_n\equiv10\cdot\underbrace{222\ldots222}_{n-2\text{ copies}}\pmod3`

Recall that a decimal integer is divisible by 3 if its digits add to a multiple of 3. The digits in
`10\cdot222\ldots222` are
`n-2` copies of 2 and one 0, so the digital sum is
`2(n-2)=2n-4`.

• If
  `n=3k` for
  `k=0,1,2,3,\ldots`, then the digital sum is
  `2(3k)-4=6k-4`, which is not divisible by 3.
• If
  `n=3k+1`, then the sum is
  `2(3k+1)-4=6k-2`, which is not divisible by 3.
• If
  `n=3k+2`, then the sum is
  `2(3k+2)-4=6k`, which is always divisble by 3.

This means that roughly 1/3 of the first
`n` numbers in this sequence are divisible by 3; among the first 100 terms, they occur for
`n=2,5,8,\ldots,95,98`, of which there are 33.

Answer 2

33

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.