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2022 AMC 10B Problem 6

Problem 6 of 25EasierAlgebraNumber Theory

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

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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=0n10k⋅10n \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 .
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Tagged: prime · factoring

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