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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 5"

(Created page with "==Problem== Define the <math>\textit{relationship}</math> between two numbers <math>a</math> and <math>b</math> to be <math>\frac{\sigma(ab)}{\sigma(a)\sigma(b)}</math> where...")
 
 
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==Solution==
 
==Solution==
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Since <math>360 = 2^3 \cdot 3^2 \cdot 5</math>, we have <math>\sigma(360) = 4 \cdot 3 \cdot 2 = 24</math>. For any number <math>n</math> such that <math>\frac{\sigma(360n)}{24\sigma(n)} = \frac{3}{4}</math>, we get <cmath>\frac{\sigma(360n)}{\sigma(n)} = 18.</cmath>
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Let <math>n = 2^a \cdot 3^b \cdot 5^c \cdot p</math> where <math>\gcd(p, 360) = 1</math>. Then the expression becomes <cmath>\frac{(a + 4)(b + 3)(c + 2)\sigma(p)}{(a + 1)(b + 1)(c + 1)\sigma(p)} = 18.</cmath>
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Substituting <math>d = a + 1</math>, <math>e = b + 1</math>, and <math>f = c + 1</math>, we get <cmath>\frac{(d + 3)(e + 2)(f + 1)}{def} = 18.</cmath>
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Try <math>d = 1</math>. Then <cmath>\frac{(1 + 3)(e + 2)(f + 1)}{e f} = \frac{4(e + 2)(f + 1)}{ef} = 18 \Rightarrow \frac{(e + 2)(f + 1)}{ef} = \frac{9}{2}.</cmath> Expanding gives <cmath>7ef - 2e - 4f - 4 = 0.</cmath>
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Using Simon's Favorite Factoring Trick, <cmath>(7e - 4)(7f - 2) = 36,</cmath> which has a solution of <math>e = 1</math>, <math>f = 2</math>.
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So <math>d = 1</math>, <math>e = 1</math>, and <math>f = 2</math>, which means <math>a = 0</math>, <math>b = 0</math>, and <math>c = 1</math>. Therefore, <math>n = 5p</math> for some <math>p</math> relatively prime to 360.
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Since <math>n \leq 100</math>, we have <math>p \leq 20</math>. The values of <math>p</math> satisfying this are <math>1, 7, 11, 13, 17, 19</math>. The sum of such integers <math>n</math> is <cmath>5 \cdot (1 + 7 + 11 + 13 + 17 + 19) = \boxed{340}.</cmath>
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~SMO_Team

Latest revision as of 21:01, 9 September 2025

Problem

Define the $\textit{relationship}$ between two numbers $a$ and $b$ to be $\frac{\sigma(ab)}{\sigma(a)\sigma(b)}$ where $\sigma(x)$ is the number of divisors of $x$. Find the sum of integers $1 \le n \le 100$ which have a relationship of $\frac{3}{4}$ with $360$.

Solution

Since $360 = 2^3 \cdot 3^2 \cdot 5$, we have $\sigma(360) = 4 \cdot 3 \cdot 2 = 24$. For any number $n$ such that $\frac{\sigma(360n)}{24\sigma(n)} = \frac{3}{4}$, we get \[\frac{\sigma(360n)}{\sigma(n)} = 18.\]

Let $n = 2^a \cdot 3^b \cdot 5^c \cdot p$ where $\gcd(p, 360) = 1$. Then the expression becomes \[\frac{(a + 4)(b + 3)(c + 2)\sigma(p)}{(a + 1)(b + 1)(c + 1)\sigma(p)} = 18.\]

Substituting $d = a + 1$, $e = b + 1$, and $f = c + 1$, we get \[\frac{(d + 3)(e + 2)(f + 1)}{def} = 18.\]

Try $d = 1$. Then \[\frac{(1 + 3)(e + 2)(f + 1)}{e f} = \frac{4(e + 2)(f + 1)}{ef} = 18 \Rightarrow \frac{(e + 2)(f + 1)}{ef} = \frac{9}{2}.\] Expanding gives \[7ef - 2e - 4f - 4 = 0.\]

Using Simon's Favorite Factoring Trick, \[(7e - 4)(7f - 2) = 36,\] which has a solution of $e = 1$, $f = 2$.

So $d = 1$, $e = 1$, and $f = 2$, which means $a = 0$, $b = 0$, and $c = 1$. Therefore, $n = 5p$ for some $p$ relatively prime to 360.

Since $n \leq 100$, we have $p \leq 20$. The values of $p$ satisfying this are $1, 7, 11, 13, 17, 19$. The sum of such integers $n$ is \[5 \cdot (1 + 7 + 11 + 13 + 17 + 19) = \boxed{340}.\]

~SMO_Team

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