2021 AMC 12A Spring Problem 25

Attempt Problem 25 of the 2021 AMC 12A Spring 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 2021 AMC 12A Spring solutions, or check the answer key.

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25.

Let d(n)d(n) denote the number of positive integers that divide n,n, including 11 and n.n. For example, d(1)=1,d(1) = 1, d(2)=2,d(2) = 2, and d(12)=6.d(12) = 6. (This function is known as the divisor function.) Let f(n)=d(n)n3. f(n) = \frac{d(n)}{\sqrt[3]{n}}.

There is a unique positive integer NN such that f(N)>f(n)f(N) \gt f(n) for all positive integers nN.n \ne N. What is the sum of the digits of N?N?

55

66

77

88

99

Answer: E
Concepts:factor countingprime factorizationoptimization
Difficulty rating: 2610
Solution:

Since f(n)=d(n)n1/3f(n) = \dfrac{d(n)}{n^{1/3}} is multiplicative, its value factors over prime powers as a product of terms e+1pe/3\dfrac{e + 1}{p^{e/3}} for each prime power pe  n.p^e\ \|\ n. We maximize each term separately.

If gp(e)=e+1pe/3,g_p(e)=\dfrac{e+1}{p^{e/3}}, then gp(e+1)gp(e)=e+2e+1p1/3.\dfrac{g_p(e+1)}{g_p(e)}=\dfrac{e+2}{e+1}p^{-1/3}. This ratio decreases with e,e, so checking where it first falls below 11 finds the unique maximum. It gives e=3e=3 for p=2,p=2, e=2e=2 for p=3,p=3, e=1e=1 for p=5,7,p=5,7, and e=0e=0 for every prime p11.p\ge11.

Hence N=233257=2520,N = 2^3\cdot 3^2\cdot 5\cdot 7 = 2520, whose digit sum is 2+5+2+0=9.2 + 5 + 2 + 0 = 9.

Thus, the correct answer is E.

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