2022 AMC 10B Problem 17

Attempt Problem 17 of the 2022 AMC 10B below, then check your answer against the professionally curated solution 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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17.

One of the following numbers is not divisible by any prime number less than 10.10. Which is it?

2606−1 2^{606}-1

2606+1 2^{606}+1

2607−1 2^{607}-1

2607+1 2^{607}+1

2607+3607 2^{607}+3^{607}

Answer: C
Concepts:factoringmodular arithmeticdivisibility
Difficulty rating: 1820
Small Hint:

For the remaining choice, test divisibility by 2,2, 3,3, 5,5, and 77

Big Hint:

Eliminate four choices by finding a small prime divisor

Solution:

Use the fact that an−bna^n-b^n is divisible by a−b.a-b.

Choice A is 2606−1=4303−1,2^{606}-1=4^{303}-1, which is divisible by 4−1=3.4-1=3.

Choice B is 2606+1=4303−(−1)303,2^{606}+1=4^{303}-(-1)^{303}, which is divisible by 4−(−1)=5.4-(-1)=5.

Choice D is 2607+1.2^{607}+1. Since 2606−12^{606}-1 is divisible by 3,3, multiplying by 22 gives 2607−22^{607}-2 divisible by 3,3, so 2607+12^{607}+1 is divisible by 3.3.

Choice E is 3607+2607=3607−(−2)607,3^{607}+2^{607}=3^{607}-(-2)^{607}, which is divisible by 3−(−2)=5.3-(-2)=5.

For choice C, 2607−12^{607}-1 is odd. Also 2607≡2(mod3),2^{607}\equiv2\pmod3, 2607≡3(mod5),2^{607}\equiv3\pmod5, and 2607≡2(mod7),2^{607}\equiv2\pmod7, so 2607−12^{607}-1 is not divisible by 3,5,3,5, or 7.7.

Thus, our answer is C .

Problem 16#16
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