Difference between revisions of "2024 AMC 10A Problems/Problem 23"
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</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: | ||
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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> | 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== | ||
− | |||
− | |||
− | |||
− | ca + b = 60 | + | <cmath>\begin{align} |
− | + | ab + c &= 100 \ | |
− | <math>\ | + | 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)== | ||
+ | |||
+ | Before we start, keep in mind that the problem is asking for the sum | ||
+ | |||
+ | 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, | ||
+ | From this observation, we conclude that the answer cannot be | ||
+ | |||
+ | Now let's test the next option, option C. | ||
+ | Option | ||
+ | |||
+ | |||
+ | |||
+ | This sum is too large. Furthermore, if all three numbers are negative, the solution still fails. For example, testing | ||
− | + | Finally, let's test the last two options: D and E. | |
+ | For option | ||
− | + | \(247 - 284 = -37\) | |
− | + | Testing values such as | |
+ | 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 | ||
+ | So, the correct answer is: | ||
+ | <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== | ==See also== |
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.
Contents
[hide]Problem
Integers , , and satisfy , , and . What is ?
Solution 1
Subtracting the first two equations yields . Notice that both factors are integers, so could equal one of and . We consider each case separately:
For , from the second equation, we see that . Then , which is not possible as is an integer, so this case is invalid.
For , we have and , which by experimentation on the factors of has no solution, so this is also invalid.
For , we have and , which by experimentation on the factors of has no solution, so this is also invalid.
Thus, we must have , so and . Thus , so . We can simply trial and error this to find that so then . The answer is then .
~eevee9406
minor edits by Lord_Erty09
Solution 2
Adding up first two equations:
Subtracting equation 1 from equation 2:
Which implies that from
Giving us that
Therefore,
~lptoggled
Solution 3 (Guess and check)
The idea is that you could guess values for , since then and are factors of . The important thing to realize is that , , and are all negative. Then, this can be solved in a few minutes, giving the solution , which gives the answer ~andliu766
Solution 4
Note that , and the only possible pair of results that yields this is and , so .
Therefore,
~luckuso, yuvag, Technodoggo (LaTeX credits to the latter two and editing to the latter)
Solution 5
There are ordered pairs of : , , .
However, only the last ordered pair meets all three equations.
Therefore,
~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
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,
Now let's test the next option, option C.
Option
This sum is too large. Furthermore, if all three numbers are negative, the solution still fails. For example, testing
Finally, let's test the last two options: D and E.
For option
Testing values such as
Once we have this answer, we still need to verify it by testing out numbers:
Finally, we test option
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.