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 00:56, 18 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
We can cancel the
since we are multiplying them on both sides of the equation.
We have
From here, it is obvious that
-iiRishabii