Difference between revisions of "2021 AMC 12A Problems/Problem 25"
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Consider the prime factorization <cmath>n=\prod_{i=1}^{k}p_i^{e_i}.</cmath> By the Multiplication Principle, <cmath>d(n)=\prod_{i=1}^{k}(e_i+1).</cmath> Now, we rewrite <math>f(n)</math> as <cmath>f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_1+1}{p_i^{{e_i}/3}}.</cmath> As <math>f(n)>0</math> for all positive integers <math>n,</math> it follows that for all positive integers <math>a</math> and <math>b</math>, <math>f(a)>f(b)</math> if and only if <math>f(a)^3>f(b)^3.</math> So, <math>f(n)</math> is maximized if and only if <cmath>f(n)^3=\prod_{i=1}^{k}\frac{(e_1+1)^3}{p_i^{{e_i}}}</cmath> is maximized. | Consider the prime factorization <cmath>n=\prod_{i=1}^{k}p_i^{e_i}.</cmath> By the Multiplication Principle, <cmath>d(n)=\prod_{i=1}^{k}(e_i+1).</cmath> Now, we rewrite <math>f(n)</math> as <cmath>f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_1+1}{p_i^{{e_i}/3}}.</cmath> As <math>f(n)>0</math> for all positive integers <math>n,</math> it follows that for all positive integers <math>a</math> and <math>b</math>, <math>f(a)>f(b)</math> if and only if <math>f(a)^3>f(b)^3.</math> So, <math>f(n)</math> is maximized if and only if <cmath>f(n)^3=\prod_{i=1}^{k}\frac{(e_1+1)^3}{p_i^{{e_i}}}</cmath> is maximized. | ||
− | For every factor <math>\frac{(e_i+1)^3}{p_i^{e_i}}</math> with a fixed <math>p_i</math> where <math>1\leq i\leq k,</math> the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime <math>p_i=2,3,5,7,\cdots,</math> we look for the <math>e_i</math> for which <math>\frac{(e_i+1)^3}{p_i^{e_i}}</math> is a relative maximum: | + | For every factor <math>\frac{(e_i+1)^3}{p_i^{e_i}}</math> with a fixed <math>p_i,</math> where <math>1\leq i\leq k,</math> the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime <math>p_i=2,3,5,7,\cdots,</math> we look for the <math>e_i</math> for which <math>\frac{(e_i+1)^3}{p_i^{e_i}}</math> is a relative maximum: |
<cmath>\begin{array}{c|c|c|c} | <cmath>\begin{array}{c|c|c|c} | ||
\boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{(e_i+1)^3/\left(p_i^{e_i}\right)} & \textbf{max?} \\ | \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{(e_i+1)^3/\left(p_i^{e_i}\right)} & \textbf{max?} \\ |
Revision as of 17:41, 18 February 2021
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 (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 Solution by Punxsutawney Phil
https://youtube.com/watch?v=y_7s8fvMCdI
Video Solution
https://www.youtube.com/watch?v=gWaUNz0gLE0
Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving )
~ 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.