Difference between revisions of "2024 AMC 10A Problems/Problem 23"
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==Solution 5== | ==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>. | |
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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. | 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> | + | 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] | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso], megaboy6679 (formatting), Technodoggo (LaTeX optimization/clarity adjustments) |
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==See also== | ==See also== |
Revision as of 17:45, 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.
Contents
[hide]Problem
Integers , , and satisfy , , and . What is ?
Solution
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)
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.