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

(Solution 5)
m (Solution)
Line 12: Line 12:
 
</math>
 
</math>
  
==Solution==
+
==Solution 1==
 
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:
 
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:
  

Revision as of 21:53, 10 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 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 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

\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)

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

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