Difference between revisions of "2019 AMC 8 Problems/Problem 17"
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==Solution 3== | ==Solution 3== | ||
− | Rewriting the numerator and the denominator, we get <math>\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}</math>. We can simplify by canceling 99! on both sides, leaving us with: <math>\frac{100 \cdot 98!}{2 | + | Rewriting the numerator and the denominator, we get <math>\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}</math>. We can simplify by canceling 99! on both sides, leaving us with: <math>\frac{100 \cdot 98!}{2 \cdot 99!}</math> We rewrite <math>99!</math> as <math>99 \cdot 98!</math> and cancel <math>98!</math>, which gets <math>\boxed{\frac{50}{99}}</math>. Answer B. |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2019|num-b=16|num-a=18}} | {{AMC8 box|year=2019|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:37, 3 February 2020
Contents
[hide]Problem 17
What is the value of the product
Solution 1
We rewrite:
The middle terms cancel, leaving us with
~phoenixfire
Solution 2
If you calculate the first few values of the equation, all of the values tend to , but are not equal to it. The answer closest to but not equal to it is .~heeeeeeeheeeee
Solution 3
Rewriting the numerator and the denominator, we get . We can simplify by canceling 99! on both sides, leaving us with: We rewrite as and cancel , which gets . Answer B.
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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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