Difference between revisions of "2017 AMC 10A Problems/Problem 1"
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| − | ==Problem== | + | == Problem == |
What is the value of <math>(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)</math>? | What is the value of <math>(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)</math>? | ||
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| − | ==Solution 1== | + | == 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 | + | a_2 = 3 \cdot 2 + 1 = 7.\\ |
| − | a_3 = 7 | + | a_3 = 7 \cdot 2 + 1 = 15.\\ |
| − | a_4 = 15 | + | a_4 = 15 \cdot 2 + 1 = 31.\\ |
| − | a_5 = 31 | + | a_5 = 31 \cdot 2 + 1 = 63.\\ |
| − | a_6 = 63 | + | a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127} |
\end{split}</cmath> | \end{split}</cmath> | ||
| − | ==Solution 2== | + | Minor LaTeX edits by fasterthanlight |
| + | |||
| + | == 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 <math>\boxed{\textbf{(C)}\ 127}</math>. | 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 <math>\boxed{\textbf{(C)}\ 127}</math>. | ||
| − | ==Solution 3== | + | == Solution 3 == |
Working our way from the innermost parenthesis outwards and directly computing, we have <math>\boxed{\textbf{(C) } 127}</math>. | Working our way from the innermost parenthesis outwards and directly computing, we have <math>\boxed{\textbf{(C) } 127}</math>. | ||
| − | ==Solution 4== | + | == 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>. | 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 | + | ==Video Solution== |
| − | + | https://youtu.be/str7kmcRMY8 | |
| + | |||
| + | https://youtu.be/kA6W8SwjitA | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | == 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 17:09, 16 June 2020
Problem
What is the value of 
?
Solution 1
Notice this is the term 
in a recursive sequence, defined recursively as 
Thus:
![\[\begin{split} a_2 = 3 \cdot 2 + 1 = 7.\\ a_3 = 7 \cdot 2 + 1 = 15.\\ a_4 = 15 \cdot 2 + 1 = 31.\\ a_5 = 31 \cdot 2 + 1 = 63.\\ a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127} \end{split}\]](http://latex.artofproblemsolving.com/b/5/d/b5d5c0648d386af95131a5512af608dc33d8e429.png)
Minor LaTeX edits by fasterthanlight
Solution 2
Starting to compute the inner expressions, we see the results are 
. This is always 
less than a power of 
. The only admissible answer choice by this rule is thus 
.
Solution 3
Working our way from the innermost parenthesis outwards and directly computing, we have 
.
Solution 4
If you distribute this you get a sum of the powers of . The largest power of in the series is 
, so the sum is .
Video Solution
~savannahsolver
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. 
