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 1== | + | === 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>. | ||
Revision as of 19:04, 27 September 2018
Contents
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. 
