2017 AMC 10B Problem 14

Attempt Problem 14 of the 2017 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 2017 AMC 10B solutions, or check the answer key.

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14.

An integer NN is selected at random in the range 1N20201\leq N \leq 2020 . What is the probability that the remainder when N16N^{16} is divided by 55 is 1?1?

15 \dfrac{1}{5}

25 \dfrac{2}{5}

35 \dfrac{3}{5}

45 \dfrac{4}{5}

1 1

Answer: D
Concepts:Fermat’s Little Theoremmodular exponentiationbasic probability
Difficulty rating: 1370
Solution:

By Fermat's Little Theorem, N41(mod5)N^4\equiv1\pmod5 whenever NN is not divisible by 5.5. Therefore, N16(N4)41(mod5).N^{16}\equiv(N^4)^4\equiv1\pmod5.

There are 2020/5=4042020/5=404 multiples of 5,5, so there are 2020404=16162020-404=1616 allowable values of N.N.

A multiple of 55 has N160(mod5),N^{16}\equiv0\pmod5, so no other values work. Thus the probability is 16162020=45.\dfrac{1616}{2020}=\dfrac45.

Thus, the correct answer is D .

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