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

Revision as of 21:31, 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 $(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(D) }276}$ .

~eevee9406

Solution 2

$(a+c)(b+1)=187$ $b+1=\pm 11,\pm 17$ $b=-12,10,-18,16$ $(a-c)(b-1)=13$ $b-1=±1,±13$ $b=0,2,-12,14$ $\rightarrow b=-12$ Which implies that $a+c=-17$ Therefore, $ab+ba+ac=100+87+60-(a+b+c)$ $=\boxed{\text{(D) }276}$ ~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

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

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

@@ Line 26: / Line 26: @@
 ==Solution 2==
 <cmath>(a+c)(b+1)=187</cmath>
-<cmath>b+1=±11,±17</cmath>
+<cmath>b+1=\pm 11,\pm 17</cmath>
 <cmath>b=-12,10,-18,16</cmath>
 <cmath>(a-c)(b-1)=13</cmath>

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