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 1

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 eacfdusituiosucioufoucio fndsaharucjifire all oif yiurhe[oifdusifh suigafhfiuaicjkgjnturh case separately: 3 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 $ca=52$, which by experimentation on the factors of $58$ has no solution, so this is also invalid.

For $b=14$, we have $14c+a=87$ 333333and $ca=46$, 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 anithium

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

\begin{align} ab + c &= 100 \\ bc + a &= 87 \\ ca + b &= 60 \end{align}

\[(1) + (2) \implies  ab + c +bc + a = (a+c)(b+1)=187\implies b+1=\pm 11,\pm 17\]

\[(1) - (2) \implies ab + c - bc - a = (a-c)(b-1)=13\implies b-1=\pm 1,\pm 13\]

Note that $(b+1)-(b-1)=2$, and the only possible pair of results that yields this is $b-1=-13$ and $b+1=-11$, so $a+c=-17$.

Therefore,

\[ab+ba+ac=ab + c +bc + a + ca + b -(a+b+c) = (1)+(2)+(3) - (a+b+c) = 100+87+60-(a+b+c)=\boxed{\textbf{(D) }276}.\] ~luckuso, yuvag, Technodoggo (LaTeX credits to the latter two and editing to the latter)

Solution 5

\begin{align} ab + c &= 100 \\ bc + a &= 87 \\ ca + b &= 60 \end{align}

(1)(2)ab+cbca=(ac)(b1)=13(2)(3)bc+acab=(ba)(c1)=27(3)(1)ca+babc=(cb)(a1)=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 (formatting), Technodoggo (LaTeX optimization/clarity adjustments)

Solution 6 (Elimination)

Before we start, keep in mind that the problem is asking for the sum ab+bc+ac. This is nothing but 100+87+60abc, or 247(a+b+c).

To solve the problem, we systematically test the options using elimination:

Let's first check options A and B, since they only happen when a,b, and c sum to 35 or 0. We begin by testing three positive values, but none satisfy the equation when there is a plus sign. For example, (12,8,4) satisfies ab+c=100, but does not satisfy bc+a=87, or ac+b=60. If a+b+c=0, then not all of the numbers can be positive or negative, so this would not work. From this observation, we conclude that the answer cannot be A or B.

Now let's test the next option, option C. Option C states ab+bc+ca=258. If true, then:

a+b+c=11

This sum is too large. Furthermore, if all three numbers are negative, the solution still fails. For example, testing (4,5,2) confirms the equation is not satisfied, as we get results that are too small. Thus, we eliminate option C.

Finally, let's test the last two options: D and E. For option E, the sum a+b+c would be:

247284=37

Testing values such as (11,12,14), the resulting sums ab+c, bc+a, and ac+b are far too large to satisfy the equation. Therefore, E is also eliminated.

Once we have this answer, we still need to verify it by testing out numbers: Finally, we test option D. Using ab+bc+ca=276, we get that a+b+c=29. Also note that a, b, and c all have to be different, because the sums from the three equations are all different. We want to get the three closest values of a, b, and c such that they are all different, and the sum a+b+c=29. The values (9,12,8) are the closest three numbers. When we try them, they satisfy all three equations. So, the correct answer is: $ab+ba+ac= -9*-12+-12*-8+-8*-9 = \boxed{\textbf{(D) }276}.$

~pimathmonkey

Video Solution by Power Solve

https://www.youtube.com/watch?v=LNYzBhf3Ke0

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

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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