Difference between revisions of "2001 AMC 10 Problems/Problem 12"
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+ | ==Video Solution by Daily Dose of Math== | ||
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+ | https://youtu.be/Ce2zWT2A0sU?si=FxjhpB2Tq0vHcl9B | ||
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+ | ~Thesmartgreekmathdude | ||
== See Also == | == See Also == |
Latest revision as of 20:42, 15 July 2024
Contents
[hide]Problem
Suppose that is the product of three consecutive integers and that is divisible by . Which of the following is not necessarily a divisor of ?
Solutions
Solution 1
Whenever is the product of three consecutive integers, is divisible by , meaning it is divisible by .
It also mentions that it is divisible by , so the number is definitely divisible by all the factors of .
In our answer choices, the one that is not a factor of is .
Solution 2
We can look for counterexamples. For example, letting , we see that is not divisible by 28, so is our answer.
Solution 3(elimination)
No matter what 3 integers you choose, one of them has to be even, so since , and it has 7 and 2 as a divisor, answer B is out. Now, if it wasn't divisible by 3, it could be A or C(,and )m so it must be divisible by 3. Therefore, it is either D or E. Since we eliminated 6, if it was E, it would be not divisible by 6(), but it is not, so the answer is .
~idk12345678
Video Solution by Daily Dose of Math
https://youtu.be/Ce2zWT2A0sU?si=FxjhpB2Tq0vHcl9B
~Thesmartgreekmathdude
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.