2021 AMC 12A Problems/Problem 25
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 each independent factor with a fixed where the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. For each prime with we look for the for which is a relative maximum:
Finally, the positive integer we seek is The sum of its digits is
Actually, once we notice that is a factor of we can conclude that the sum of the digits of must be a multiple of Only choice is possible.
~MRENTHUSIASM
Solution 2 (Fast)
Using the answer choices to our advantage, we can show that must be divisible by 9 without explicitly computing , by exploiting the following fact:
Claim: If is not divisible by 3, then .
Proof: Since is a multiplicative function, we have and . Then
Note that the values and do not have to be explicitly computed; we only need the fact that which is easy to show by hand.
The above claim automatically implies is a multiple of 9: if was not divisible by 9, then which is a contradiction, and if was divisible by 3 and not 9, then , also a contradiction. Then the sum of digits of must be a multiple of 9, so only choice works.
-scrabbler94
Video Solutions
https://www.youtube.com/watch?v=gWaUNz0gLE0 (by Dedekind Cuts)
https://www.youtube.com/watch?v=Sv4gj1vMjOs (by Aaron He)
https://youtube.com/watch?v=y_7s8fvMCdI (by Punxsutawney Phil)
https://youtu.be/6P-0ZHAaC_A (by OmegaLearn)
~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.