Difference between revisions of "2024 AMC 10A Problems/Problem 23"
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− | + | ==Solution 2== | |
+ | <math>(a+c)(b+1)=187</math> | ||
+ | <math>b+1=±11,±17</math> | ||
+ | <math>b=-12,10,-18,16</math> | ||
+ | <math>(a-c)(b-1)=13</math> | ||
+ | <math>b-1=±1,±13</math> | ||
+ | <math>b=0,2,-12,14</math> | ||
+ | <math>\rightarrow b=-12</math> | ||
+ | Which implies that <math>a+c=-17</math> | ||
+ | Therefore, ab+ba+ac=100+87+60-(a+b+c) | ||
+ | <math>=\boxed{\text{(B) }276}</math> | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}} | ||
{{AMC12 box|year=2024|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2024|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:32, 8 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
Solution 2
Which implies that Therefore, ab+ba+ac=100+87+60-(a+b+c)
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.