Difference between revisions of "2020 AMC 8 Problems/Problem 12"
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<cmath>5\cdot2\cdot1\cdot9! = N!</cmath> | <cmath>5\cdot2\cdot1\cdot9! = N!</cmath> | ||
We see that <math>5\cdot2\cdot1\cdot9! = 10\cdot9! = 10! = N!</math>. So <math>N = \boxed{10} = \textbf{(A)}10</math>. | We see that <math>5\cdot2\cdot1\cdot9! = 10\cdot9! = 10! = N!</math>. So <math>N = \boxed{10} = \textbf{(A)}10</math>. | ||
− | + | ==Solution 6== | |
+ | We note that <math>5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!</math> we can actually get 120*9!= 12*N! which then you just get to your conclusion 10! which is equal to answer choice <math>N=\textbf{(A)}10</math>. | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/9k59v-Fr3aE | https://youtu.be/9k59v-Fr3aE |
Revision as of 18:03, 19 November 2020
For a positive integer , the factorial notation represents the product of the integers from to . What value of satisfies the following equation?
Contents
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 's, since we are multiplying them on both sides of the equation.
We have
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
Solution 5
We see that . Notice that , so: We see that . So .
Solution 6
We note that we can actually get 120*9!= 12*N! which then you just get to your conclusion 10! which is equal to answer choice .
Video Solution
~savannahsolver
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 |
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