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

m (Protected "2024 AMC 10A Problems/Problem 23" ([Edit=Allow only administrators] (expires 04:59, 8 November 2024 (UTC)) [Move=Allow only administrators] (expires 04:59, 8 November 2024 (UTC))) [cascading])
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==Problem==
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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>?
  
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<math>
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\textbf{(A) }212 \qquad
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\textbf{(B) }247 \qquad
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\textbf{(C) }258 \qquad
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\textbf{(D) }276 \qquad
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\textbf{(E) }284 \qquad
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</math>
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==Solution==
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Subtracting the first two equations yields <math>(a-c)(b-1)=13</math>. Notice that both factors are integers, so <math>b-1</math> could equal one of <math>13,1,-1,-13</math> and <math>b=14,2,0,-12</math>. We consider each case separately:
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For <math>b=0</math>, from the second equation, we see that <math>a=87</math>. Then <math>80c=60</math>, which is not possible as <math>c</math> is an integer, so this case is invalid.
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For <math>b=2</math>, we have <math>2c+a=87</math> and <math>ca=58</math>, which by experimentation on the factors of <math>58</math> has no solution, so this is also invalid.
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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.
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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>.
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~eevee9406
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==See also==
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{{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}}
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{{MAA Notice}}

Revision as of 16:22, 8 November 2024

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

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