2024 AMC 10A Problems/Problem 23

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

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

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

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

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

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

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

There are $3$ ordered pairs of $(a,b,c)$ : $(5,14,4)$ , $(-3,-12,-3)$ , $(-9,-12,-8)$

However, only the last ordered pair meets all three equations.

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

~luckuso ~megaboy6679 for formats

2024 AMC 10A (Problems • Answer Key • Resources)
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2024 AMC 12A (Problems • Answer Key • Resources)
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