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kumpel [21]
3 years ago
5

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 $3$?
Mathematics
2 answers:
asambeis [7]3 years ago
8 0

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.

balandron [24]3 years ago
7 0

The nth number in the sequence can be expressed as

a_n=10^{n+1}+\displaystyle2\sum_{i=2}^n10^i+21

Extracting the nth 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.

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