2016 AMC 10A Problems/Problem 1

Problem

What is the value of $\dfrac{11!-10!}{9!}$ ?

$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$

Solution 1

We can use subtraction of fractions to get $\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.$

Solution 2

Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}$ .

Solution 3

$\dfrac{11!-10!}{9!}$ consider 10 as n $\dfrac{(n+1)!-n!}{(n-1)!}$ simpify $\dfrac{(n+1)n!-(-1)n!}{(n-1)!}$ = $\dfrac{n(n!)}{(n-1)!}$ = $\dfrac{n(n(n-1)!)}{(n-1)!}$ = $\dfrac{n(n)(1)}{(1}$ = $\dfrac{n^2}{1}$ subsitute n as 10 again $\dfrac{10^2}{1}$

answer is $10^2$ which is 100

Solution 4

We are given the equation $\frac{11!-10!}{9!}$

This is equivalent to $\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\frac{(11-1)(10!)}{9!}$ , which equals $10 \cdot 10$ .

Therefore, the answer is $10^2$ = $\boxed{\textbf{(B)}~100}$ .

~TheGoldenRetriever

Solution 5 (This is a joke)

Let $I_n := \int_0^\infty x^n e^{-x} \, dx = n!$ (the Gamma–integral).

Then $\frac{11! - 10!}{9!} = \frac{I_{11} - I_{10}}{I_9} = \frac{\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx}{I_9}.$ Integration by parts on the numerator with $ u = x^{11} - x^{10} $, $ dv = e^{-x} \, dx $ (so $ du = (11x^{10} - 10x^9) \, dx $, $ v = -e^{-x} $) gives $\int_0^\infty e^{-x} (x^{11} - x^{10}) \, dx = \left[ -e^{-x} (x^{11} - x^{10}) \right]_0^\infty + \int_0^\infty e^{-x} (11x^{10} - 10x^9) \, dx = 11 I_{10} - 10 I_9,$ since the boundary term vanishes.

Hence $\frac{I_{11} - I_{10}}{I_9} = \frac{11 I_{10}}{I_9} - 10.$ Do another integration by parts to relate $ I_{10} $ and $ I_9 $: $I_{10} = \int_0^\infty x^{10} e^{-x} \, dx = \left[ -x^{10} e^{-x} \right]_0^\infty + 10 \int_0^\infty x^9 e^{-x} \, dx = 10 I_9.$ Therefore $\frac{I_{11} - I_{10}}{I_9} = 11 \cdot 10 - 10 = 100.$

$\boxed{(B) 100}$ .

~Pinotation

~Minorly Edited by OffBrandCab

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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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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