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

(Solution 4)
(Solution 4)
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The idea is that you could guess values for <math>c</math>, since then <math>a</math> and <math>b</math> are factors of <math>100 - c</math>. The important thing to realize is that <math>a</math>, <math>b</math>, and <math>c</math> are all negative. Then, this can be solved in a few minutes, giving the solution <math>(-9, -12, -8)</math>, which gives the answer <math>\boxed{\textbf{(D)} 276}</math>
 
The idea is that you could guess values for <math>c</math>, since then <math>a</math> and <math>b</math> are factors of <math>100 - c</math>. The important thing to realize is that <math>a</math>, <math>b</math>, and <math>c</math> are all negative. Then, this can be solved in a few minutes, giving the solution <math>(-9, -12, -8)</math>, which gives the answer <math>\boxed{\textbf{(D)} 276}</math>
 
~andliu766
 
~andliu766
 +
  
 
==Solution 4==
 
==Solution 4==
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~"latexified" by yuvag
 
~"latexified" by yuvag
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 +
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==Solution 5==
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<math>ab + c = 100 \text{ } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}</math>
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<math>bc + a = 87  \text{            } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}</math>
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<math>ca + b = 60  \text{            } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}</math> 
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<math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} =  ab + c -bc - a =(a-c)(b-1)=13</math>
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<math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} =  bc + a -ca - b =(b-a)(c-1)=27</math>
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<math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} =  ca + b -ab - c =(c-b)(a-1)=-40</math>
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so 3 groups of possible answers
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(A)  a = 5, b = 14, c = 4
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(B)  a = -3, b = -12, c = -3
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(C)  a = -9, b = -12 , c = -8
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only group (C) meet all three equation.
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Therefore, <math>ab+ba+ac= -9*-12+-12*-8+-8*-9 = \boxed{\textbf{(D) }276}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
 
==See also==
 
==See also==

Revision as of 11:16, 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

$ab + c = 100 \text{ } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}$

$bc + a = 87  \text{            } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}$

$ca + b = 60  \text{            } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}$


$\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} + \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} =  ab + c +bc + a = (a+c)(b+1)=187$

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


$\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} =  ab + c -bc - a =(a-c)(b-1)=13$

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


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

$\implies a+c=-17$


Therefore, $ab+ba+ac=ab + c +bc + a + ca + b -(a+b+c) = \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}+\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}+\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} - (a+b+c) = 100+87+60-(a+b+c)$ $=\boxed{\textbf{(D) }276}$ ~luckuso

~"latexified" by yuvag


Solution 5

$ab + c = 100 \text{ } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}$

$bc + a = 87  \text{            } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}$

$ca + b = 60  \text{            } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}$

$\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} =  ab + c -bc - a =(a-c)(b-1)=13$

$\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} =  bc + a -ca - b =(b-a)(c-1)=27$

$\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} =  ca + b -ab - c =(c-b)(a-1)=-40$

so 3 groups of possible answers

(A) a = 5, b = 14, c = 4

(B) a = -3, b = -12, c = -3

(C) a = -9, b = -12 , c = -8

only group (C) meet all three equation.

Therefore, $ab+ba+ac= -9*-12+-12*-8+-8*-9 = \boxed{\textbf{(D) }276}$

~luckuso

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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