2023 AMC 12B Problems/Problem 24

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 (Guesstimation)

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 $\boxed{\textbf{(C) 3}}

==Solution 2==

Denote by$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)\nu_2 (x) $for$ x \in \left\{ a, b, c, d \right\}$.

Equation (5) implies$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)\nu_3 (x) $for$ x \in \left\{ a, b, c, d \right\}$.

Equation (2) implies$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)\nu_5 (x) $for$ x \in \left\{ a, b, c, d \right\}$.

Equation (5) implies$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)\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$ (Error compiling LaTeX. Unknown error_msg)\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, <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)

==Solution 3 (GCD/LCM Comparison)==

We are given that$ (Error compiling LaTeX. Unknown error_msg)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,$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)a $,$ b $,$ c $,$ d$.

Thus, the only possible nontrivial common factor among$ (Error compiling LaTeX. Unknown error_msg)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)

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2023 AMC 12B Problems/Problem 24

Contents

Problem

Solution 1 (Guesstimation)

Video Solution

See Also