Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution: The sequence is not an arithmetic sequence)
(→‎Solution)
Line 6: Line 6:
  
  
−
==Solution==
+
==Solution 1==
  
−
Notice this is the term <math>a_6</math> in a recursive sequence, defined recursively as <math>a_1 = 3, a_n = 2a_(n-1) + 1.</math> Thus:
+
Notice this is the term <math>a_6</math> in a recursive sequence, defined recursively as <math>a_1 = 3, a_n = 2a_{n-1} + 1.</math> Thus:
 
<cmath>\begin{split}
 
<cmath>\begin{split}
 
a_2 = 3*2 + 1 = 7.\\
 
a_2 = 3*2 + 1 = 7.\\
Line 14: Line 14:
 
a_4 = 15*2 + 1 = 31.\\
 
a_4 = 15*2 + 1 = 31.\\
 
a_5 = 31*2 + 1 = 63.\\
 
a_5 = 31*2 + 1 = 63.\\
−
a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127.}
+
a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127}
 
\end{split}</cmath>
 
\end{split}</cmath>
 +
 +
==Solution 2==
 +
Starting to compute the inner expressions, we see the results are <math>1, 3, 7, 15, \ldots</math>. This is always <math>1</math> less than a power of <math>2</math>. The only admissible answer choice by this rule is thus \boxed{\textbf{(C)}\ 127}.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:52, 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}.

See Also

2017 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_1&oldid=83062"