Difference between revisions of "2023 AMC 12B Problems/Problem 24"
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<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== | + | ==Solution 1 (Guesstimation)== |
| − | Denote by <math>\nu_p (x)< | + | 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}} |
| + | |||
| + | ==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)< | + | 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< | + | 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< | + | 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< | + | 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< | + | 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< | + | 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)< | + | 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< | + | 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< | + | 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< | + | 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< | + | 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< | + | 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)< | + | 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< | + | 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< | + | Equation (2) implies </math>\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3<math>. |
| − | Thus, <math>\nu_5 (a) = 3< | + | Thus, </math>\nu_5 (a) = 3<math>. |
| − | From Equations (5)-(7), we have either <math>\nu_5 (b) \leq 1< | + | 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< | + | Equation (1) implies </math>\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7<math>. |
| − | Thus, for <math>\nu_5 (b)< | + | 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 64: | Line 68: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| − | ==Solution | + | ==Solution 3 (GCD/LCM Comparison)== |
| − | We are given that <math>abcd = 2^6 \cdot 3^9 \cdot 5^7< | + | 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> |
| − | Comparing this product with <math>abcd< | + | 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> |
| − | This additional factor <math>2^1 \cdot 3^4 \cdot 5^2< | + | 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< | + | 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< | + | 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< | + | Thus, the only possible nontrivial common factor among </math>a<math>, </math>b<math>, </math>c<math>, </math>d$ is 3. |
Therefore, | Therefore, | ||
Revision as of 01:40, 19 September 2025
Problem
Suppose that 
, 
, 
and 
are positive integers satisfying all of the following relations.
![\[abcd=2^6\cdot 3^9\cdot 5^7\]](http://latex.artofproblemsolving.com/2/9/e/29e86e43b8f649b005defff6a2fad55137f1f63c.png)
![\[\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3\]](http://latex.artofproblemsolving.com/9/6/e/96e948a17f61e59d4b8d3a22eb5fb3ae138313e4.png)
![\[\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3\]](http://latex.artofproblemsolving.com/d/8/9/d894077385e2b22f41c6d1e8cc018dae3b148cae.png)
![\[\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3\]](http://latex.artofproblemsolving.com/5/6/d/56df1c066d690462e5674b637bcb874a62b53721.png)
![\[\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2\]](http://latex.artofproblemsolving.com/1/0/4/104e9a748bef440670593f6a39cd7c7ad329ca90.png)
![\[\text{lcm}(b,d)=2^2\cdot 3^3\cdot 5^2\]](http://latex.artofproblemsolving.com/8/d/7/8d7e6c3cd5cfc60253742c496aa5d3610ec049a9.png)
![\[\text{lcm}(c,d)=2^2\cdot 3^3\cdot 5^2\]](http://latex.artofproblemsolving.com/c/8/3/c837d2c0e335f03ada1a0a87f3058689ad6dcebc.png)
What is 
?
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)
p
x$.
We index Equations given in this problem from (1) to (7).
First, we compute$ (Error compiling LaTeX. Unknown error_msg)\nu_2 (x)
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
\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3
\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2
\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
\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)
\left( 1, 0 \right)$.
Second, we compute$ (Error compiling LaTeX. Unknown error_msg)\nu_3 (x)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
\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3
\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3\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\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)\left( 2, 1 \right)$.
Third, we compute$ (Error compiling LaTeX. Unknown error_msg)\nu_5 (x)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\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3
\nu_5 (a) = 3$.
From Equations (5)-(7), we have either$ (Error compiling LaTeX. Unknown error_msg)\nu_5 (b) \leq 1
\nu_5 (c) = \nu_5 (d) = 2
\nu_5 (b) = 2\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
\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
\[
\mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) = (2^3 \cdot 3^2 \cdot 5^3) \cdot (2^2 \cdot 3^3 \cdot 5^2) = 2^{5} \cdot 3^{5} \cdot 5^{5}.
\]
abcd
\[
abcd = 2^6 \cdot 3^9 \cdot 5^7 = \mathrm{lcm}(a,b) \cdot \mathrm{lcm}(c,d) \cdot 2^1 \cdot 3^4 \cdot 5^2.
\]
2^1 \cdot 3^4 \cdot 5^2
abcd$, which are their common divisors.
The key observation is that the only prime with a leftover exponent greater than 3 is 3 (specifically,$ (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)abcd$.
Thus, the only possible nontrivial common factor among$ (Error compiling LaTeX. Unknown error_msg)abcd$ is 3.
Therefore,
\[ \gcd(a,b,c,d) = \boxed {3}. \]
~Mewoooow
Video Solution
~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. 
