2020 AMC 8 Problems/Problem 12: Difference between revisions
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<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 | ~ junaidmansuri | ||
==Solution 3 (Non-rigorous)== | |||
We can see that the answers B through E have the factor 11, but there is no 11 in <math>5!\cdot9!</math>. Therefore, the answer must be the only answer without a <math>11</math> factor, <math>A</math>. | |||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=11|num-a=13}} | {{AMC8 box|year=2020|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:42, 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?
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
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.
~ junaidmansuri
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,
.
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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