Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 10A Problems/Problem 23"
m
Line 1: Line 1:
 +
{{duplicate|[[2024 AMC 10A Problems/Problem 23|2024 AMC 10A #23]] and [[2024 AMC 12A Problems/Problem 17|2024 AMC 12A #17]]}}
 +
 
==Problem==
 
==Problem==
 
Integers <math>a</math>, <math>b</math>, and <math>c</math> satisfy <math>ab + c = 100</math>, <math>bc + a = 87</math>, and <math>ca + b = 60</math>. What is <math>ab + bc + ca</math>?
 
Integers <math>a</math>, <math>b</math>, and <math>c</math> satisfy <math>ab + c = 100</math>, <math>bc + a = 87</math>, and <math>ca + b = 60</math>. What is <math>ab + bc + ca</math>?
Line 19: Line 21:
 
For <math>b=14</math>, we have <math>14c+a=87</math> and <math>ca=46</math>, which by experimentation on the factors of <math>46</math> has no solution, so this is also invalid.
 
For <math>b=14</math>, we have <math>14c+a=87</math> and <math>ca=46</math>, which by experimentation on the factors of <math>46</math> has no solution, so this is also invalid.
  
−
Thus, we must have <math>b=-12</math>, so <math>a=12c+87</math> and <math>ca=72</math>. Thus <math>c(12c+87)=72</math>, so <math>c(4c+29)=24</math>. We can simply trial and error this to find that <math>c=-8</math>, so then <math>a=-9</math>. The answer is then <math>(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(B) }276}</math>.
+
Thus, we must have <math>b=-12</math>, so <math>a=12c+87</math> and <math>ca=72</math>. Thus <math>c(12c+87)=72</math>, so <math>c(4c+29)=24</math>. We can simply trial and error this to find that <math>c=-8</math> so then <math>a=-9</math>. The answer is then <math>(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(B) }276}</math>.
  
 
~eevee9406
 
~eevee9406
Line 25: Line 27:
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}}
 
{{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}}
 +
{{AMC10 box|year=2024|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:42, 8 November 2024

The following problem is from both the 2024 AMC 10A #23 and 2024 AMC 12A #17, so both problems redirect to this page.

Problem

Integers $a$, $b$, and $c$ satisfy $ab + c = 100$, $bc + a = 87$, and $ca + b = 60$. What is $ab + bc + ca$?

$\textbf{(A) }212 \qquad \textbf{(B) }247 \qquad \textbf{(C) }258 \qquad \textbf{(D) }276 \qquad \textbf{(E) }284 \qquad$

Solution

Subtracting the first two equations yields $(a-c)(b-1)=13$. Notice that both factors are integers, so $b-1$ could equal one of $13,1,-1,-13$ and $b=14,2,0,-12$. We consider each case separately:

For $b=0$, from the second equation, we see that $a=87$. Then $80c=60$, which is not possible as $c$ is an integer, so this case is invalid.

For $b=2$, we have $2c+a=87$ and $ca=58$, which by experimentation on the factors of $58$ has no solution, so this is also invalid.

For $b=14$, we have $14c+a=87$ and $ca=46$, which by experimentation on the factors of $46$ has no solution, so this is also invalid.

Thus, we must have $b=-12$, so $a=12c+87$ and $ca=72$. Thus $c(12c+87)=72$, so $c(4c+29)=24$. We can simply trial and error this to find that $c=-8$ so then $a=-9$. The answer is then $(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(B) }276}$.

~eevee9406

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 10 Problems and Solutions
2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_23&oldid=232054"