Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12B Problems/Problem 24"

(Created page with "Suppose that <math>a,b,c,</math> and <math>d</math> are positive integers satisfying all of the following relations. What is <math>gcd(a,b,c,d)?</math>")
 
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
−
Suppose that <math>a,b,c,</math> and <math>d</math> are positive integers satisfying all of the following relations.
+
==Problem==
−
What is <math>gcd(a,b,c,d)?</math>
+
Suppose that <math>a</math>, <math>b</math>, <math>c</math> and <math>d</math> are positive integers satisfying all of the following relations.
 +
 
 +
<cmath>abcd=2^6\cdot 3^9\cdot 5^7</cmath>
 +
<cmath>\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3</cmath>
 +
<cmath>\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3</cmath>
 +
<cmath>\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3</cmath>
 +
<cmath>\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2</cmath>
 +
<cmath>\text{lcm}(b,d)=2^2\cdot 3^3\cdot 5^2</cmath>
 +
<cmath>\text{lcm}(c,d)=2^2\cdot 3^3\cdot 5^2</cmath>
 +
 
 +
What is <math>\text{gcd}(a,b,c,d)</math>?
 +
 
 +
<math>\textbf{(A)}~30\qquad\textbf{(B)}~45\qquad\textbf{(C)}~3\qquad\textbf{(D)}~15\qquad\textbf{(E)}~6</math>
 +
 
 +
==Solution==
 +
 
 +
Denote by <math>\nu_p (x)</math> the number of prime factor <math>p</math> in number <math>x</math>.
 +
 
 +
We index Equations given in this problem from (1) to (7).
 +
 
 +
 
 +
First, we compute <math>\nu_2 (x)</math> for <math>x \in \left\{ a, b, c, d \right\}</math>.
 +
 
 +
Equation (5) implies <math>\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1</math>.
 +
Equation (2) implies <math>\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3</math>.
 +
Equation (6) implies <math>\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2</math>.
 +
Equation (1) implies <math>\nu_2 (a) + \nu_2 (b) + \nu_2 (c) + \nu_2 (d) = 6</math>.
 +
 
 +
Therefore, all above jointly imply <math>\nu_2 (a) = 3</math>, <math>\nu_2 (d) = 2</math>, and <math>\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)</math> or <math>\left( 1, 0 \right)</math>.
 +
 
 +
 
 +
Second, we compute <math>\nu_3 (x)</math> for <math>x \in \left\{ a, b, c, d \right\}</math>.
 +
 
 +
Equation (2) implies <math>\max \left\{ \nu_3 (a), \nu_3 (b) \right\} = 2</math>.
 +
Equation (3) implies <math>\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3</math>.
 +
Equation (4) implies <math>\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3</math>.
 +
Equation (1) implies <math>\nu_3 (a) + \nu_3 (b) + \nu_3 (c) + \nu_3 (d) = 9</math>.
 +
 
 +
Therefore, all above jointly imply <math>\nu_3 (c) = 3</math>, <math>\nu_3 (d) = 3</math>, and <math>\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)</math> or <math>\left( 2, 1 \right)</math>.
 +
 
 +
Third, we compute <math>\nu_5 (x)</math> for <math>x \in \left\{ a, b, c, d \right\}</math>.
 +
 
 +
 
 +
Equation (5) implies <math>\max \left\{ \nu_5 (b), \nu_5 (c) \right\} = 2</math>.
 +
Equation (2) implies <math>\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3</math>.
 +
Thus, <math>\nu_5 (a) = 3</math>.
 +
 
 +
From Equations (5)-(7), we have either <math>\nu_5 (b) \leq 1</math> and <math>\nu_5 (c) = \nu_5 (d) = 2</math>, or <math>\nu_5 (b) = 2</math> and <math>\max \left\{ \nu_5 (c), \nu_5 (d) \right\} = 2</math>.
 +
 
 +
Equation (1) implies <math>\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7</math>.
 +
Thus, for <math>\nu_5 (b)</math>, <math>\nu_5 (c)</math>, <math>\nu_5 (d)</math>, there must be two 2s and one 0.
 +
 
 +
Therefore,
 +
<cmath>
 +
\begin{align*}
 +
{\rm gcd} (a,b,c,d)
 +
& = \Pi_{p \in \{ 2, 3, 5\}} p^{\min\{ \nu_p (a), \nu_p(b) , \nu_p (c), \nu_p(d) \}} \\
 +
& = 2^0 \cdot 3^1 \cdot 5^0 \\
 +
& = \boxed{\textbf{(C) 3}}.
 +
\end{align*}
 +
</cmath>
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/RtkZTYrpE-w
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2023|ab=B|num-b=23|num-a=25}}
 +
{{MAA Notice}}

Latest revision as of 01:57, 20 November 2023

Problem

Suppose that $a$, $b$, $c$ and $d$ are positive integers satisfying all of the following relations.

\[abcd=2^6\cdot 3^9\cdot 5^7\] \[\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3\] \[\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2\] \[\text{lcm}(b,d)=2^2\cdot 3^3\cdot 5^2\] \[\text{lcm}(c,d)=2^2\cdot 3^3\cdot 5^2\]

What is $\text{gcd}(a,b,c,d)$?

$\textbf{(A)}~30\qquad\textbf{(B)}~45\qquad\textbf{(C)}~3\qquad\textbf{(D)}~15\qquad\textbf{(E)}~6$

Solution

Denote by $\nu_p (x)$ the number of prime factor $p$ in number $x$.

We index Equations given in this problem from (1) to (7).


First, we compute $\nu_2 (x)$ for $x \in \left\{ a, b, c, d \right\}$.

Equation (5) implies $\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1$. Equation (2) implies $\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3$. Equation (6) implies $\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2$. Equation (1) implies $\nu_2 (a) + \nu_2 (b) + \nu_2 (c) + \nu_2 (d) = 6$.

Therefore, all above jointly imply $\nu_2 (a) = 3$, $\nu_2 (d) = 2$, and $\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)$ or $\left( 1, 0 \right)$.


Second, we compute $\nu_3 (x)$ for $x \in \left\{ a, b, c, d \right\}$.

Equation (2) implies $\max \left\{ \nu_3 (a), \nu_3 (b) \right\} = 2$. Equation (3) implies $\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3$. Equation (4) implies $\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3$. Equation (1) implies $\nu_3 (a) + \nu_3 (b) + \nu_3 (c) + \nu_3 (d) = 9$.

Therefore, all above jointly imply $\nu_3 (c) = 3$, $\nu_3 (d) = 3$, and $\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)$ or $\left( 2, 1 \right)$.

Third, we compute $\nu_5 (x)$ for $x \in \left\{ a, b, c, d \right\}$.


Equation (5) implies $\max \left\{ \nu_5 (b), \nu_5 (c) \right\} = 2$. Equation (2) implies $\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3$. Thus, $\nu_5 (a) = 3$.

From Equations (5)-(7), we have either $\nu_5 (b) \leq 1$ and $\nu_5 (c) = \nu_5 (d) = 2$, or $\nu_5 (b) = 2$ and $\max \left\{ \nu_5 (c), \nu_5 (d) \right\} = 2$.

Equation (1) implies $\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7$. Thus, for $\nu_5 (b)$, $\nu_5 (c)$, $\nu_5 (d)$, there must be two 2s and one 0.

Therefore, \begin{align*} {\rm gcd} (a,b,c,d) & = \Pi_{p \in \{ 2, 3, 5\}} p^{\min\{ \nu_p (a), \nu_p(b) , \nu_p (c), \nu_p(d) \}} \\ & = 2^0 \cdot 3^1 \cdot 5^0 \\ & = \boxed{\textbf{(C) 3}}. \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/RtkZTYrpE-w

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_24&oldid=204753"