2022 AMC 10A Problem 19

Attempt Problem 19 of the 2022 AMC 10A 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 10A solutions, or check the answer key.

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19.

Let LnL_n denote the least common multiple of the numbers 1,1, 2,2, 3,3, ,\ldots, n,n, and let hh be the unique positive integer such that 11+12+13++117=hL17\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{17} = \dfrac{h}{L_{17}} What is the remainder when hh is divided by 17?17?

11

33

55

77

99

Answer: C
Concepts:modular arithmeticleast common multipleprime factorization
Difficulty rating: 2150
Small Hint:

Work modulo 1717 after multiplying by the least common multiple

Big Hint:

All terms except the one from 117\frac{1}{17} vanish modulo 1717

Solution:

Multiplying the harmonic sum by L17,L_{17}, we get h=i=117L17i.h=\sum_{i=1}^{17}\frac{L_{17}}{i}.

For 1i16,1\le i\le16, the term L17i\frac{L_{17}}{i} is still divisible by 17,17, so these terms contribute 0(mod17).0\pmod{17}.

Thus hL1717(mod17).h\equiv \frac{L_{17}}{17}\pmod{17}. The least common multiple L17L_{17} contains the prime-power factors 16,9,5,7,11,13,17,16,9,5,7,11,13,17, so L1717169571113(mod17). \begin{aligned} \frac{L_{17}}{17} &\equiv 16\cdot9\cdot5\cdot7\cdot11\cdot13 \\ &\pmod{17}. \end{aligned}

Reducing modulo 17,17, this is (1)9571113(-1)\cdot9\cdot5\cdot7\cdot11\cdot13 5(mod17).\equiv5\pmod{17}.

Thus, C is the correct answer.

Problem 18#18
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