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2023 AMC 12B Problems/Problem 24: Difference between revisions

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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.


==Solution==
<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 1==


Denote by <math>\nu_p (x)</math> the number of prime factor <math>p</math> in number <math>x</math>.
Denote by <math>\nu_p (x)</math> the number of prime factor <math>p</math> in number <math>x</math>.
Line 51: Line 63:


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
==Solution 2 (GCD/LCM Comparison)==
We are given that <imath>abcd = 2^6 \cdot 3^9 \cdot 5^7</imath>, and several LCM relations involving pairs of these numbers. Notice that
<imath>
\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}.
</imath>
Comparing this product with <imath>abcd</imath>, we see that
<imath>
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.
</imath>
This additional factor <imath>2^1 \cdot 3^4 \cdot 5^2</imath> must be accounted for by the overlaps among <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath>, which are their common divisors.
The key observation is that the only prime with a leftover exponent greater than 3 is 3 (specifically, <imath>3^4</imath>), 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 <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath>.
Thus, the only possible nontrivial common factor among <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath> is 3.
Therefore,
<imath>\gcd(a,b,c,d) = \boxed {3}.</imath>
~Mewoooow
~Latex fix by megacleverstarfish15
==Video Solutions==
https://youtu.be/RtkZTYrpE-w
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
https://www.youtube.com/watch?v=kky_f4JK7y8&feature=youtu.be
~megacleverstarfish15
==See Also==
{{AMC12 box|year=2023|ab=B|num-b=23|num-a=25}}
{{MAA Notice}}

Latest revision as of 18:41, 11 November 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 ~Latex fix by megacleverstarfish15

Video Solutions

https://youtu.be/RtkZTYrpE-w

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

https://www.youtube.com/watch?v=kky_f4JK7y8&feature=youtu.be

~megacleverstarfish15

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.

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