2022 AMC 10B Problem 6

Below is the professionally curated solution for Problem 6 of the 2022 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

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Concepts:primefactoring

Difficulty rating: 1140

6.

How many of the first ten numbers of the sequence 121,11211,1112111,121, 11211, 1112111, \ldots are prime numbers?

0 0

1 1

2 2

3 3

4 4

Solution:

We claim that none of these numbers can ever be prime.

We prove this claim by noticing that the nnth number is k=02n10k+10n=k=0n10k+\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k + k=n2n10k=k=0n10k+k=0n10k10n \sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n =(10n+1)(k=0n10k).= (10^n+1)(\sum_{k=0}^{n} 10^k). This shows that the number can be written as the product of two numbers greater than 1,1, so there are no primes.

Thus, the answer is A .

Problem 6 in Other Years