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Difference between revisions of "2024 AMC 10A Problems/Problem 23"

(→‎Solution 4)
(→‎Solution 4)
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==Solution 4==
 
==Solution 4==
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ab + c = 100   (1)
+
<math>ab + c = 100 \textcircled{    1}</math>
 
      
 
      
−
bc + a = 87    (2)  
+
bc + a = 87    \textcircled{    2}<math>  
  
−
ca + b = 60    (3)    
+
ca + b = 60    \textcircled{    3}</math>    
 
   
 
   
−
(1) + (2) =  ab + c +bc + a = (a+c)(b+1)=187  
+
<math>\textcircled{1} + \textcircled{2} =  ab + c +bc + a = (a+c)(b+1)=187</math>
−
<cmath>b+1=\pm 11,\pm 17</cmath>
 
  
−
(1) - (2) =  ab + c -bc - a =(a-c)(b-1)=13
+
<math>b+1=\pm 11,\pm 17</math>
−
<cmath>b-1=\pm 1,\pm 13</cmath>
 
  
−
the only possible pair that has difference of 2 is b-1=-13 , b+1= -11 , then b=-12 ,
+
<math>\textcircled{1} - \textcircled{2} =  ab + c -bc - a =(a-c)(b-1)=13</math>
−
Which implies that <math>a+c=-17</math>
+
 
−
Therefore, <math>ab+ba+ac=ab + c +bc + a + ca + b -(a+b+c) = (1)+(2)+(3) -(a+b+c) = 100+87+60-(a+b+c)</math>
+
<math>b-1=\pm 1, \pm 13</math>
−
<cmath>=\boxed{\text{(D) }276}</cmath>
+
 
 +
The only possible pair that has difference of <math>2</math> is <math>b-1=-13</math>, <math>b+1= -11</math>, then <math>b=-12</math>
 +
 
 +
<math>\implies a+c=-17</math>
 +
 
 +
Therefore, <math>ab+ba+ac=ab + c +bc + a + ca + b -(a+b+c) = \textcircled{1}+\textcircled{2}+\textcircled{3} - (a+b+c) = 100+87+60-(a+b+c)</math>
 +
<math>=\boxed{\textbf{(D) }276}</math>
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 +
 +
~Slightly "latexified" by yuvag
  
 
==See also==
 
==See also==

Revision as of 10:33, 9 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 $87c=60$, 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 $(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(D) }276}$.

~eevee9406 

minor edits by Lord_Erty09

Solution 2

Adding up first two equations: \[(a+c)(b+1)=187\] \[b+1=\pm 11,\pm 17\] \[b=-12,10,-18,16\]

Subtracting equation 1 from equation 2: \[(a-c)(b-1)=13\] \[b-1=\pm 1,\pm 13\] \[b=0,2,-12,14\]

\[\Rightarrow b=-12\]

Which implies that $a+c=-17$ from $(a+c)(b+1)=187$

Giving us that $a+b+c=-29$

Therefore, $ab+bc+ac=100+87+60-(a+b+c)=\boxed{\text{(D) }276}$

~lptoggled

Solution 3 (Guess and check)

The idea is that you could guess values for $c$, since then $a$ and $b$ are factors of $100 - c$. The important thing to realize is that $a$, $b$, and $c$ are all negative. Then, this can be solved in a few minutes, giving the solution $(-9, -12, -8)$, which gives the answer $\boxed{\textbf{(D)} 276}$ ~andliu766

Solution 4

$ab + c = 100 \textcircled{ 1}$

bc + a = 87 \textcircled{ 2}$ca + b = 60 \textcircled{ 3}$

$\textcircled{1} + \textcircled{2} = ab + c +bc + a = (a+c)(b+1)=187$

$b+1=\pm 11,\pm 17$

$\textcircled{1} - \textcircled{2} = ab + c -bc - a =(a-c)(b-1)=13$

$b-1=\pm 1, \pm 13$

The only possible pair that has difference of $2$ is $b-1=-13$, $b+1= -11$, then $b=-12$

$\implies a+c=-17$

Therefore, $ab+ba+ac=ab + c +bc + a + ca + b -(a+b+c) = \textcircled{1}+\textcircled{2}+\textcircled{3} - (a+b+c) = 100+87+60-(a+b+c)$ $=\boxed{\textbf{(D) }276}$ ~luckuso

~Slightly "latexified" by yuvag

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 12A (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 12 Problems and Solutions

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