Difference between revisions of "2016 AMC 10A Problems/Problem 1"
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<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math> | <math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | <math>\frac{11!-10!}{9!}=\frac{11\cdot10!-10!}{9!}=\frac{100\cdot9!}{9!}=100</math> | |
+ | |||
+ | |||
+ | <math>\boxed{\textbf{(B)}~100}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.</cmath> | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Factoring out <math>9!</math> gives <math>\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/VIt6LnkV4_w | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/CrS7oHDrvP8 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}} | {{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}} | ||
+ | {{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:16, 16 June 2020
Problem
What is the value of ?
Solution 1
Solution 2
We can use subtraction of fractions to get
Solution 3
Factoring out gives .
Video Solution
~IceMatrix
~savannahsolver
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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 AMC 10 Problems and Solutions |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by First Problem |
Followed by Problem 2 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.