Difference between revisions of "2017 AMC 10A Problems/Problem 1"
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==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>. |
==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 11:54, 12 February 2017
Problem
What is the value of 
?
Solution 1
Notice this is the term 
in a recursive sequence, defined recursively as 
Thus:
![\[\begin{split} a_2 = 3*2 + 1 = 7.\\ a_3 = 7 *2 + 1 = 15.\\ a_4 = 15*2 + 1 = 31.\\ a_5 = 31*2 + 1 = 63.\\ a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127} \end{split}\]](http://latex.artofproblemsolving.com/2/4/f/24f2ed2ffa26184a9cd739e4ebd30f4a9f3c9286.png)
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 .
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 | ||
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