Difference between revisions of "2020 AMC 8 Problems/Problem 12"
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Solution 1 | Solution 1 | ||
| − | Notice that <math>5!</math> = <math>2*3*4*5,</math> and we can combine the numbers to | + | Notice that <math>5!</math> = <math>2*3*4*5,</math> and we can combine the numbers to create a larger factorial. To turn <math>9!</math> into <math>10!,</math> we need to multiply <math>9!</math> by <math>2*5,</math> which equals to <math>10!.</math> |
| + | |||
| + | Therefore, we have | ||
<cmath>10!*12=12*N!.</cmath> | <cmath>10!*12=12*N!.</cmath> | ||
| − | + | We can cancel the <math>12s,</math> since we are multiplying them on both sides of the equation. | |
| + | |||
| + | We have | ||
<cmath>10!=N!.</cmath> | <cmath>10!=N!.</cmath> | ||
| − | + | From here, it is obvious that <math>N=10(A).</math> | |
-iiRishabii | -iiRishabii | ||
Revision as of 23:56, 17 November 2020
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 
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
