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

(→‎Solution 2 (GCD/LCM Comparison))
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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
 +
 
\[
 
\[
 
\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}.
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Comparing this product with <math>abcd</math>, we see that
 
Comparing this product with <math>abcd</math>, 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.
 
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.
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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 4 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>.
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Therefore,
 
Therefore,
 +
 
\[
 
\[
−
\gcd(a,b,c,d) = \boxed{3}.
+
\gcd(a,b,c,d) = \boxed {3}.
 
\]
 
\]
  

Latest revision as of 11:11, 6 August 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

\[ \mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) = (2^3 \cdot 3^2 \cdot 5^3) \cdot (2^2 \cdot 3^3 \cdot 5^2) = 2^{5} \cdot 3^{5} \cdot 5^{5}. \]

Comparing this product with $abcd$, we see that

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

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

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