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

(→‎Solution 4)
(→‎Solution 6 (Elimination))
 
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</math>
 
</math>
  
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==Solution==
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==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:
  
Line 48: Line 48:
 
The idea is that you could guess values for <math>c</math>, since then <math>a</math> and <math>b</math> are factors of <math>100 - c</math>. The important thing to realize is that <math>a</math>, <math>b</math>, and <math>c</math> are all negative. Then, this can be solved in a few minutes, giving the solution <math>(-9, -12, -8)</math>, which gives the answer <math>\boxed{\textbf{(D)} 276}</math>
 
The idea is that you could guess values for <math>c</math>, since then <math>a</math> and <math>b</math> are factors of <math>100 - c</math>. The important thing to realize is that <math>a</math>, <math>b</math>, and <math>c</math> are all negative. Then, this can be solved in a few minutes, giving the solution <math>(-9, -12, -8)</math>, which gives the answer <math>\boxed{\textbf{(D)} 276}</math>
 
~andliu766
 
~andliu766
 +
  
 
==Solution 4==
 
==Solution 4==
−
<math>ab + c = 100 \text{ } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}</math>
 
−
   
 
−
<math>bc + a = 87  \text{            } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}</math>
 
  
−
<math>ca + b = 60 \text{           } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}</math>
+
<cmath>\begin{align}
 +
ab + c &= 100 \\
 +
bc + a &= 87 \\
 +
ca + b &= 60
 +
\end{align}</cmath>
 +
 
 +
<cmath>(1) + (2) \implies  ab + c +bc + a = (a+c)(b+1)=187\implies b+1=\pm 11,\pm 17</cmath>
 +
 
 +
<cmath>(1) - (2) \implies ab + c - bc - a = (a-c)(b-1)=13\implies b-1=\pm 1,\pm 13</cmath>
 +
 
 +
Note that <math>(b+1)-(b-1)=2</math>, and the only possible pair of results that yields this is <math>b-1=-13</math> and <math>b+1=-11</math>, so <math>a+c=-17</math>.
 +
 
 +
Therefore,
 +
 
 +
<cmath>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}.</cmath>
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso], yuvag, Technodoggo (LaTeX credits to the latter two and editing to the latter)
 +
 
 +
==Solution 5==
 +
<cmath>\begin{align}
 +
ab + c &= 100 \\
 +
bc + a &= 87 \\
 +
ca + b &= 60
 +
\end{align}</cmath>
 +
 
 +
\begin{align*}
 +
(1) - (2) \implies ab + c -bc - a &=(a-c)(b-1)=13 \\
 +
(2) - (3) \implies bc + a -ca - b &=(b-a)(c-1)=27 \\
 +
(3) - (1) \implies ca + b -ab - c &=(c-b)(a-1)=-40
 +
\end{align*}
 +
 
 +
There are <math>3</math> ordered pairs of <math>(a,b,c)</math>: <math>(5,14,4)</math>, <math>(-3,-12,-3)</math>, <math>(-9,-12,-8)</math>.
 +
 
 +
However, only the last ordered pair meets all three equations.
 +
 
 +
Therefore, <math>ab+ba+ac= -9*-12+-12*-8+-8*-9 = \boxed{\textbf{(D) }276}.</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso], megaboy6679 (formatting), Technodoggo (LaTeX optimization/clarity adjustments)
  
−
+
==Solution 6 (Elimination)==
−
 
−
<math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} + \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} = ab + c +bc + a = (a+c)(b+1)=187</math>
 
  
−
<math>b+1=\pm 11,\pm 17</math>
+
Before we start, keep in mind that the problem is asking for the sum \(ab+bc+ac\). This is nothing but \(100+87+60-a-b-c\), 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 \( \textbf{A} \) or \( \textbf{B} \).
  
−
<math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} =  ab + c -bc - a =(a-c)(b-1)=13</math>
+
Now let's test the next option, option C.
 +
Option \( \textbf{C} \) states \( ab + bc + ca = 258 \). If true, then:
  
−
<math>b-1=\pm 1, \pm 13</math>
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\(a + b + c = -11\)
  
 +
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 \( \textbf{C} \).
  
 +
Finally, let's test the last two options: D and E.
 +
For option \( \textbf{E} \), the sum \( a + b + c \) would be:
  
−
The only possible pair that has difference of <math>2</math> is <math>b-1=-13</math>, <math>b+1= -11</math>, then <math>b=-12</math>
+
\(247 - 284 = -37\)
  
−
<math>\implies a+c=-17</math>
+
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, \( \textbf{E} \) is also eliminated.
  
 +
Once we have this answer, we still need to verify it by testing out numbers:
 +
Finally, we test option \( \textbf{D} \). 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:
 +
<math>ab+ba+ac= -9*-12+-12*-8+-8*-9 = \boxed{\textbf{(D) }276}.</math>
  
 +
~pimathmonkey
  
−
Therefore, <math>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)</math>
+
== Video Solution by Power Solve ==
−
<math>=\boxed{\textbf{(D) }276}</math>
+
https://www.youtube.com/watch?v=LNYzBhf3Ke0
−
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
  
−
~"latexified" by yuvag
+
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=6SQ74nt3ynw
  
 
==See also==
 
==See also==

Latest revision as of 20:28, 17 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}

\begin{align*} (1) - (2) \implies ab + c -bc - a &=(a-c)(b-1)=13 \\ (2) - (3) \implies bc + a -ca - b &=(b-a)(c-1)=27 \\ (3) - (1) \implies ca + b -ab - c &=(c-b)(a-1)=-40 \end{align*}

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 \(100+87+60-a-b-c\), 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 \( \textbf{A} \) or \( \textbf{B} \).

Now let's test the next option, option C. Option \( \textbf{C} \) 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 \( \textbf{C} \).

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

\(247 - 284 = -37\)

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, \( \textbf{E} \) is also eliminated.

Once we have this answer, we still need to verify it by testing out numbers: Finally, we test option \( \textbf{D} \). 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 (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

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