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

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https://www.youtube.com/watch?v=dQw4w9WgXcQ
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{{duplicate|[[2024 AMC 10A Problems/Problem 23|2024 AMC 10A #23]] and [[2024 AMC 12A Problems/Problem 17|2024 AMC 12A #17]]}}
    ^
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see answer here
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==Problem==
 +
Integers <math>a</math>, <math>b</math>, and <math>c</math> satisfy <math>ab + c = 100</math>, <math>bc + a = 87</math>, and <math>ca + b = 60</math>. What is <math>ab + bc + ca</math>?
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<math>
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\textbf{(A) }212 \qquad
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\textbf{(B) }247 \qquad
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\textbf{(C) }258 \qquad
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\textbf{(D) }276 \qquad
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\textbf{(E) }284 \qquad
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</math>
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==Solution 1==
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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:
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3
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For <math>b=0</math>, from the second equation, we see that <math>a=87</math>. Then <math>87c=60</math>, which is not possible as <math>c</math> is an integer, so this case is invalid.
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For <math>b=2</math>, we have <math>2c+a=87</math> and <math>ca=52</math>, which by experimentation on the factors of <math>58</math> has no solution, so this is also invalid.
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For <math>b=14</math>, we have <math>14c+a=87</math> and <math>ca=46</math>, which by experimentation on the factors of <math>46</math> has no solution, so this is also invalid.
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Thus, we must have <math>b=-12</math>, so <math>a=12c+87</math> and <math>ca=72</math>. Thus <math>c(12c+87)=72</math>, so <math>c(4c+29)=24</math>. We can simply trial and error this to find that <math>c=-8</math> so then <math>a=-9</math>. The answer is then <math>(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(D) }276}</math>.
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~eevee9406
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==Solution 2==
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Adding up first two equations: <cmath>(a+c)(b+1)=187</cmath>
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<cmath>b+1=\pm 11,\pm 17</cmath>
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<cmath>b=-12,10,-18,16</cmath>
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Subtracting equation 1 from equation 2: <cmath>(a-c)(b-1)=13</cmath>
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<cmath>b-1=\pm 1,\pm 13</cmath>
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<cmath>b=0,2,-12,14</cmath>
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<cmath>\Rightarrow b=-12</cmath>
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Which implies that <math>a+c=-17</math> from <math>(a+c)(b+1)=187</math>
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 +
Giving us that <math>a+b+c=-29</math>
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Therefore, <math>ab+bc+ac=100+87+60-(a+b+c)=\boxed{\text{(D) }276}</math>
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~lptoggled
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== Solution 3 (Guess and check) ==
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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>
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~andliu766
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 +
 
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==Solution 4==
 +
 
 +
<cmath>\begin{align}
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ab + c &= 100 \\
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bc + a &= 87 \\
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ca + b &= 60
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\end{align}</cmath>
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<cmath>(1) + (2) \implies  ab + c +bc + a = (a+c)(b+1)=187\implies b+1=\pm 11,\pm 17</cmath>
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<cmath>(1) - (2) \implies ab + c - bc - a = (a-c)(b-1)=13\implies b-1=\pm 1,\pm 13</cmath>
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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>.
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Therefore,
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<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>
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso], yuvag, Technodoggo (LaTeX credits to the latter two and editing to the latter)
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 +
==Solution 5==
 +
<cmath>\begin{align}
 +
ab + c &= 100 \\
 +
bc + a &= 87 \\
 +
ca + b &= 60
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\end{align}</cmath>
 +
 
 +
\begin{align*}
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(1) - (2) \implies ab + c -bc - a &=(a-c)(b-1)=13 \\
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(2) - (3) \implies bc + a -ca - b &=(b-a)(c-1)=27 \\
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(3) - (1) \implies ca + b -ab - c &=(c-b)(a-1)=-40
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\end{align*}
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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>
 +
 
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist 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.
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From this observation, we conclude that the answer cannot be \( \textbf{A} \) or \( \textbf{B} \).
 +
 
 +
Now let's test the next option, option C.
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Option \( \textbf{C} \) states \( ab + bc + ca = 258 \). If true, then:
 +
 
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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 \( \textbf{C} \).
 +
 
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Finally, let's test the last two options: D and E.
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For option \( \textbf{E} \), the sum \( a + b + c \) would be:
 +
 
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\(247 - 284 = -37\)
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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. 
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Therefore, \( \textbf{E} \) is also eliminated.
 +
 
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Once we have this answer, we still need to verify it by testing out numbers:
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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.
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So, the correct answer is:
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<math>ab+ba+ac= -9*-12+-12*-8+-8*-9 = \boxed{\textbf{(D) }276}.</math>
 +
 
 +
~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==
 +
{{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}}
 +
{{AMC12 box|year=2024|ab=A|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Latest revision as of 14:31, 1 December 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: 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 $14c+a=87$ 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

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 (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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