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

Answer:

  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.