Difference between revisions of "2021 AMC 12A Problems/Problem 25"
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<math>\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9</math> | <math>\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9</math> | ||
− | + | ==Solution 1== | |
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Consider the prime factorization <cmath>n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.</cmath> By the Multiplication Principle, <cmath>d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).</cmath> Now, we rewrite <math>f(n)</math> as <cmath>f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1)}{{p_1}^{{e_1}/3}{p_2}^{{e_2}/3}{p_3}^{{e_3}/3}\cdots{p_k}^{{e_k}/3}}=\left(\frac{e_1+1}{{p_1}^{{e_1}/3}}\right)\left(\frac{e_2+1}{{p_2}^{{e_2}/3}}\right)\left(\frac{e_3+1}{{p_3}^{{e_3}/3}}\right)\cdots\left(\frac{e_k+1}{{p_k}^{{e_k}/3}}\right).</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=\left(\frac{(e_1+1)^3}{{p_1}^{e_1}}\right)\left(\frac{(e_2+1)^3}{{p_2}^{e_2}}\right)\left(\frac{(e_3+1)^3}{{p_3}^{e_3}}\right)\cdots\left(\frac{(e_k+1)^3}{{p_k}^{e_k}}\right)</cmath> is maximized. | Consider the prime factorization <cmath>n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.</cmath> By the Multiplication Principle, <cmath>d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).</cmath> Now, we rewrite <math>f(n)</math> as <cmath>f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1)}{{p_1}^{{e_1}/3}{p_2}^{{e_2}/3}{p_3}^{{e_3}/3}\cdots{p_k}^{{e_k}/3}}=\left(\frac{e_1+1}{{p_1}^{{e_1}/3}}\right)\left(\frac{e_2+1}{{p_2}^{{e_2}/3}}\right)\left(\frac{e_3+1}{{p_3}^{{e_3}/3}}\right)\cdots\left(\frac{e_k+1}{{p_k}^{{e_k}/3}}\right).</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=\left(\frac{(e_1+1)^3}{{p_1}^{e_1}}\right)\left(\frac{(e_2+1)^3}{{p_2}^{e_2}}\right)\left(\frac{(e_3+1)^3}{{p_3}^{e_3}}\right)\cdots\left(\frac{(e_k+1)^3}{{p_k}^{e_k}}\right)</cmath> is maximized. | ||
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~MRENTHUSIASM | ~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. <cmath>\frac{27}{9}d(k)^3 > \frac{8}{3}d(k)^3 > d(k)^3</cmath> so <math>f(9k) > f(3k) > f(k)</math> and since <math>\boxed{\textbf{(E) }9}</math> is the only possible answer choice, it is the answer. | 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. <cmath>\frac{27}{9}d(k)^3 > \frac{8}{3}d(k)^3 > d(k)^3</cmath> so <math>f(9k) > f(3k) > f(k)</math> and since <math>\boxed{\textbf{(E) }9}</math> is the only possible answer choice, it is the answer. | ||
Revision as of 17:59, 12 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
~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) | |
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 |
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