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

(Solution 5: Guessing Strategies for Lazy People)
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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>.
 
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>.
 
==Solution 5: Guessing Strategies for Lazy People==
 
"I'm too lazy, so I'll try to guess this," says the lazy person. Well then this is the solution for you! (note that in general however, these guessing strategies are just meant to be confidence boosters for your numerical thinking). If you look up a probability table for each of the answer choices on the AMC, you will get <math>\boxed{\text{C}}</math>.
 
{{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 

Revision as of 18:46, 28 July 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 $\boxed{\textbf{(C)}\ 127}$.