Difference between revisions of "2001 AMC 10 Problems/Problem 12"
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<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42 </math> | <math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42 </math> | ||
− | == Solution == | + | == Solution 1== |
Whenever <math> n </math> is the product of three consecutive integers, <math> n </math> is divisible by <math> 3! </math>, meaning it is divisible by <math> 6 </math>. | Whenever <math> n </math> is the product of three consecutive integers, <math> n </math> is divisible by <math> 3! </math>, meaning it is divisible by <math> 6 </math>. | ||
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It also mentions that it is divisible by <math> 7 </math>, so the number is definitely divisible by all the factors of <math> 42 </math>. | It also mentions that it is divisible by <math> 7 </math>, so the number is definitely divisible by all the factors of <math> 42 </math>. | ||
− | In our answer choices, the one that is not a factor of <math> 42 </math> is <math> \boxed{\textbf{(D)}\ 28} </math>. | + | In our answer choices, the one that is not a factor of <math> 42 </math> is <math> \boxed{\textbf{(D)}\ 28} </math>. |
== Solution 2 == | == Solution 2 == |
Revision as of 18:04, 27 September 2018
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 ?
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 (D) is our answer.
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.