Difference between revisions of "2020 AMC 8 Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
<math>5!\cdot 9!=12\cdot N!</math><br><math>120\cdot 9!=12\cdot N!</math><br><math>12\cdot 10\cdot 9!=12\cdot N!</math><br><math>12 \cdot 10!=12\cdot N!</math><br><math>N=10 \implies\boxed{\textbf{(A) }10}</math>.<br> | <math>5!\cdot 9!=12\cdot N!</math><br><math>120\cdot 9!=12\cdot N!</math><br><math>12\cdot 10\cdot 9!=12\cdot N!</math><br><math>12 \cdot 10!=12\cdot N!</math><br><math>N=10 \implies\boxed{\textbf{(A) }10}</math>.<br> | ||
| − | ~ junaidmansuri | + | ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri] |
==Solution 3 (Non-rigorous)== | ==Solution 3 (Non-rigorous)== | ||
Revision as of 18:34, 18 November 2020
For a positive integer 
, the factorial notation 
represents the product of the integers from to 
. What value of 
satisfies the following equation? ![\[5!\cdot 9!=12\cdot N!\]](http://latex.artofproblemsolving.com/7/a/a/7aaf22ceff79c5a5063175866cf2cba2e4a57f9b.png)
Solution 1
Notice that 
= 
and we can combine the numbers to create a larger factorial. To turn 
into 
we need to multiply by 
which equals to 
Therefore, we have
![\[10!*12=12*N!.\]](http://latex.artofproblemsolving.com/b/5/7/b57d51cc68f940f63a1dde3a63bbc125429d2bfc.png)
We can cancel the 
's, since we are multiplying them on both sides of the equation.
We have
![\[10!=N!.\]](http://latex.artofproblemsolving.com/d/d/8/dd8b5a203f6c403ffb8a87279e697ee510eefb36.png)
From here, it is obvious that 
-iiRishabii
Solution 2
.Solution 3 (Non-rigorous)
We can see that the answers B through E have the factor 11, but there is no 11 in 
. Therefore, the answer must be the only answer without a 
factor, 
.
~Windigo
Solution 4
Notice that 
. We are also told that 
from where it is obvious that 
.
-franzliszt
See also
| 2020 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
