Difference between revisions of "2017 AMC 10A Problems/Problem 1"

Revision as of 20:20, 8 February 2017

Problem

What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$ ?

$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$

Solution 1

Notice this is the term $a_6$ in a recursive sequence, defined recursively as $a_1 = 3, a_n = 2a_{n-1} + 1.$ Thus: $\begin{split} a_2 = 3*2 + 1 = 7.\\ a_3 = 7 *2 + 1 = 15.\\ a_4 = 15*2 + 1 = 31.\\ a_5 = 31*2 + 1 = 63.\\ a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127} \end{split}$

Solution 2

Starting to compute the inner expressions, we see the results are $1, 3, 7, 15, \ldots$ . This is always $1$ less than a power of $2$ . The only admissible answer choice by this rule is thus $\boxed{\textbf{(C)}\ 127}$ .

Solution 3

Working our way from the innermost parenthesis outwards and directly computing, we have $\boxed{\textbf{(C) } 127}$ .

Solution 4

If you distribute this you get a sum of the powers of $2$ . The largest power of $2$ in the series is $64$ , so the sum is $\boxed{\textbf{(C)}127}$

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 23: / Line 23: @@
 Working our way from the innermost parenthesis outwards and directly computing, we have <math>\boxed{\textbf{(C) } 127}</math>.
+==Solution 4==
+If you distribute this you get a sum of the powers of <math>2</math>.  The largest power of <math>2</math> in the series is <math>64</math>, so the sum is <math>\boxed{\textbf{(C)}127}</math>
 ==See Also==
 {{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}}
 {{MAA Notice}}

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