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

Revision as of 11:09, 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 4 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)

@@ Line 76: / Line 76: @@
 \]
-This additional factor <math>2^1 \cdot 3^4 \cdot 5^2</math> must be accounted for by the overlaps among <math>a, b, c, 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 4 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, b, c, 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, b, c, d</math> 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,

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