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"
 
(3 intermediate revisions by the same user not shown)
Line 14: Line 14:
 
<math>\textbf{(A)}~30\qquad\textbf{(B)}~45\qquad\textbf{(C)}~3\qquad\textbf{(D)}~15\qquad\textbf{(E)}~6</math>
 
<math>\textbf{(A)}~30\qquad\textbf{(B)}~45\qquad\textbf{(C)}~3\qquad\textbf{(D)}~15\qquad\textbf{(E)}~6</math>
  
−
==Solution 1 (Guesstimation)==
+
==Solution 1==
  
−
The real answer to the problem lies between the conversion of \( \text{lcm}(a,b) \times \text{lcm}(c,d) \) to \( \text{gcd}(a,b,c,d) \). We assume \( \text{lcm}(a,b,c,d) = \text{lcm}(a,b) \times \text{lcm}(b,c) \). Then, because \( \text{lcm}(a,b) = 2^3 \times 3^2 \times 5^3 \) and \( \text{lcm}(c,d) = 2^2 \times 3^3 \times 5^2 \), we see \( \text{lcm}(a,b,c,d)=2^6 \times 3^6 \times 5^6 \). We know \( \text{gcd}(x,y,\dots) = \frac{|x-y-\dots|}{\text{lcm}(x,y,\dots)} \). So then, substitution shows us \( \text{gcd}(a,b,c,d)=\frac{|2^6-3^6-5^6|}{2^6 \times 3^6 \times 5^6} \approx 2.23 \). We stated that the answer lies between this conversion, and the nearest answer choice is <math>\boxed{\textbf{(C) 3}}
+
Denote by <math>\nu_p (x)</math> the number of prime factor <math>p</math> in number <math>x</math>.
−
 
 
−
==Solution 2==
 
−
 
 
−
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).
 
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>.
+
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 (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 (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 (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>.
+
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>.
+
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>.
+
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 (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 (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 (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>.
+
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>.
+
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>.
+
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 (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>.
+
Equation (2) implies <math>\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3</math>.
−
Thus, </math>\nu_5 (a) = 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>.
+
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>.
+
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.
+
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,
 
Therefore,
Line 68: Line 64:
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
−
==Solution 3 (GCD/LCM Comparison)==
+
==Solution 2 (GCD/LCM Comparison)==
  
−
We are given that </math>abcd = 2^6 \cdot 3^9 \cdot 5^7<math>, and several LCM relations involving pairs of these numbers. Notice that
+
We are given that <math>abcd = 2^6 \cdot 3^9 \cdot 5^7</math>, and several LCM relations involving pairs of these numbers. Notice that
  
−
</math>\[
+
<math>\[
 
\mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) = (2^3 \cdot 3^2 \cdot 5^3) \cdot (2^2 \cdot 3^3 \cdot 5^2) = 2^{5} \cdot 3^{5} \cdot 5^{5}.
 
\mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) = (2^3 \cdot 3^2 \cdot 5^3) \cdot (2^2 \cdot 3^3 \cdot 5^2) = 2^{5} \cdot 3^{5} \cdot 5^{5}.
−
\]<math>
+
\]</math>
  
−
Comparing this product with </math>abcd<math>, we see that
+
Comparing this product with <math>abcd</math>, we see that
  
−
</math>\[
+
<math>\[
 
abcd = 2^6 \cdot 3^9 \cdot 5^7 = \mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) \cdot 2^1 \cdot 3^4 \cdot 5^2.
 
abcd = 2^6 \cdot 3^9 \cdot 5^7 = \mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) \cdot 2^1 \cdot 3^4 \cdot 5^2.
−
\]<math>
+
\]</math>
  
−
This additional factor </math>2^1 \cdot 3^4 \cdot 5^2<math> must be accounted for by the overlaps among </math>a<math>, </math>b<math>, </math>c<math>, </math>d<math>, which are their common divisors.
+
This additional factor <math>2^1 \cdot 3^4 \cdot 5^2</math> must be accounted for by the overlaps among <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, which are their common divisors.
  
−
The key observation is that the only prime with a leftover exponent greater than 3 is 3 (specifically, </math>3^4<math>), which suggests that all four numbers share at least one factor of 3.
+
The key observation is that the only prime with a leftover exponent greater than 3 is 3 (specifically, <math>3^4</math>), which suggests that all four numbers share at least one factor of 3.
  
−
For primes 2 and 5, the leftover exponents are too small (1 and 2, respectively) to be shared by all four numbers, since the GCD exponent must be the minimum exponent among </math>a<math>, </math>b<math>, </math>c<math>, </math>d<math>.
+
For primes 2 and 5, the leftover exponents are too small (1 and 2, respectively) to be shared by all four numbers, since the GCD exponent must be the minimum exponent among <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>.
  
−
Thus, the only possible nontrivial common factor among </math>a<math>, </math>b<math>, </math>c<math>, </math>d$ is 3.
+
Thus, the only possible nontrivial common factor among <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> is 3.
  
 
Therefore,
 
Therefore,

Latest revision as of 01:43, 19 September 2025

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 1

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)

Solution 2 (GCD/LCM Comparison)

We are given that $abcd = 2^6 \cdot 3^9 \cdot 5^7$, and several LCM relations involving pairs of these numbers. Notice that

$\[ \mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) = (2^3 \cdot 3^2 \cdot 5^3) \cdot (2^2 \cdot 3^3 \cdot 5^2) = 2^{5} \cdot 3^{5} \cdot 5^{5}. \]$ (Error compiling LaTeX. Unknown error_msg)

Comparing this product with $abcd$, we see that

$\[ abcd = 2^6 \cdot 3^9 \cdot 5^7 = \mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) \cdot 2^1 \cdot 3^4 \cdot 5^2. \]$ (Error compiling LaTeX. Unknown error_msg)

This additional factor $2^1 \cdot 3^4 \cdot 5^2$ must be accounted for by the overlaps among $a$, $b$, $c$, $d$, which are their common divisors.

The key observation is that the only prime with a leftover exponent greater than 3 is 3 (specifically, $3^4$), which suggests that all four numbers share at least one factor of 3.

For primes 2 and 5, the leftover exponents are too small (1 and 2, respectively) to be shared by all four numbers, since the GCD exponent must be the minimum exponent among $a$, $b$, $c$, $d$.

Thus, the only possible nontrivial common factor among $a$, $b$, $c$, $d$ is 3.

Therefore,

\[ \gcd(a,b,c,d) = \boxed {3}. \]

~Mewoooow

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_24&oldid=260876"