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

m (Minor LaTeX formats)
(Video Solution)
(2 intermediate revisions by 2 users not shown)
Line 15: Line 15:
 
a_5 = 31 \cdot 2 + 1 = 63.\\
 
a_5 = 31 \cdot 2 + 1 = 63.\\
 
a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127}
 
a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127}
\end{split}</cmath>.
+
\end{split}</cmath>
  
 
Minor LaTeX edits by fasterthanlight
 
Minor LaTeX edits by fasterthanlight
Line 32: Line 32:
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/str7kmcRMY8
 
https://youtu.be/str7kmcRMY8
 +
 +
https://youtu.be/kA6W8SwjitA
 +
 +
~savannahsolver
  
 
== See Also ==
 
== See Also ==

Revision as of 17:09, 16 June 2020

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: \[\begin{split} a_2 = 3 \cdot 2 + 1 = 7.\\ a_3 = 7 \cdot 2 + 1 = 15.\\ a_4 = 15 \cdot 2 + 1 = 31.\\ a_5 = 31 \cdot 2 + 1 = 63.\\ a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127} \end{split}\]

Minor LaTeX edits by fasterthanlight

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 $\boxed{\textbf{(C)}\ 127}$.

Video Solution

https://youtu.be/str7kmcRMY8

https://youtu.be/kA6W8SwjitA

~savannahsolver

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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

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