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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?

 Oct 22, 2019
 #1
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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

 

 

cool cool cool

 Oct 22, 2019

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