2021 AMC 12A Problems/Problem 25
Contents
Problem
Let denote the number of positive integers that divide , including and . For example, and . (This function is known as the divisor function.) LetThere is a unique positive integer such that for all positive integers . What is the sum of the digits of
Solution 1
Consider the prime factorization By the Multiplication Principle, Now, we rewrite as As for all positive integers it follows that for all positive integers and , if and only if So, is maximized if and only if is maximized.
For every factor with a fixed where the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime we look for the for which is a relative maximum:
Finally, the number we seek is The sum of its digits is
Actually, once we get that is a factor of we know that the sum of the digits of must be a multiple of Only choice is possible.
~MRENTHUSIASM
Solution 2
A cube root seems bad, so we should just cube it. It seems that if the number is a multiple of 3, there are only two choices. If the number is a multiple of 9, there is one choice. We can prove that for all k is indivisible by 3, f(9k) > f(3k) > f(k). The divisors of 3k contain the divisors of k and the divisors of k multiplied by 3. The divisors of 9k contain the divisors of k, the divisors of k multiplied by 3, and the divisors of k multiplied by 9. so and since is the only possible answer choice, it is the answer.
Solution 3 (Only Use If Low On Time)
We can guess that would be divisible by . Recall, for a number to be divisible by , the sum of its digits must also be divisible by . Since there's only one choice where it's divisible by , we get as our answer. ~rocketsri
Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving )
~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
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