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Difference between revisions of "2021 JMC 10 Problems/Problem 19"

(Created page with "==Problem== Two distinct divisors of <math>6^4=1296</math> are ''mutual'' if their difference divides their product. For instance, <math>(4,2)</math> is mutual as <math>(4-2)...")
 
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==Solution==
 
==Solution==
  
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Observe that <cmath>(d_1-d_2) |d_1 d_2 \implies (k-1)d_2 | k\cdot (d_2)^2 \implies (k-1) | k d_2.</cmath> Because <math>\gcd(k,k-1)=1</math>, it follows that <math>(k-1)|d_2</math>. Since <math>d_2|1296</math>, we must also have <math>(k-1)|1296</math>. Note that we cannot have <math>6|(k-1)</math>, because this will result in <math>k</math> being neither a multiple of <math>2</math> nor <math>3</math>.
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Observe that <cmath>(d_1-d_2) |d_1 d_2 \implies (k-1)d_2 | k\cdot (d_2)^2 \implies (k-1) | k d_2.</cmath> Because <math>\gcd(k,k-1)=1</math>, it follows that <math>(k-1)|d_2</math>. Since <math>d_2|1296</math>, we must also have <math>(k-1)|1296</math>. Note that we cannot have <math>6|(k-1)</math>, because this will result in <math>k</math> being neither a multiple of <math>2</math> nor <math>3</math>. Thus, we need only to check <math>(k-1)|16</math> and <math>(k-1)|81.</math> Note that <math>k</math> must be of the form <math>2^a\cdot 3^b</math> for non-negative integers <math>a</math> and <math>b,</math> so our desired answer is <math>2+3+4+9=18</math>.
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Thus, we need only to check <math>(k-1)|16</math> and <math>(k-1)|81.</math> Note that <math>k</math> must be of the form <math>2^a\cdot 3^b</math> for non-negative integers <math>a</math> and <math>b,</math> so our desired answer is <math>2+3+4+9=18</math>.
 

Latest revision as of 15:03, 1 April 2021

Problem

Two distinct divisors of $6^4=1296$ are mutual if their difference divides their product. For instance, $(4,2)$ is mutual as $(4-2)\mid 4\cdot2.$ Suppose a mutual pair $(d_1,d_2)$ exists where $d_1 = kd_2$ for a positive integer $k.$ What is the sum of all possible $k?$

$\textbf{(A) } 14 \qquad\textbf{(B) } 18 \qquad\textbf{(C) } 19 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 23$


Solution

Observe that \[(d_1-d_2) |d_1 d_2 \implies (k-1)d_2 | k\cdot (d_2)^2 \implies (k-1) | k d_2.\] Because $\gcd(k,k-1)=1$, it follows that $(k-1)|d_2$. Since $d_2|1296$, we must also have $(k-1)|1296$. Note that we cannot have $6|(k-1)$, because this will result in $k$ being neither a multiple of $2$ nor $3$. Thus, we need only to check $(k-1)|16$ and $(k-1)|81.$ Note that $k$ must be of the form $2^a\cdot 3^b$ for non-negative integers $a$ and $b,$ so our desired answer is $2+3+4+9=18$.

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