Difference between revisions of "2017 AMC 10A Problems/Problem 1"
(Created page with "==Problem== What is the value of <math>(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)</math>? <math>\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qqua...") |
(→Solution 6 (quickest)) |
||
(32 intermediate revisions by 17 users not shown) | |||
Line 1: | Line 1: | ||
− | ==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>? | ||
<math>\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729</math> | <math>\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729</math> | ||
+ | |||
+ | == 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: | ||
+ | <cmath>\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}</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Working our way from the innermost parenthesis outwards and directly computing, we have <math>\boxed{\textbf{(C) } 127}</math>. | ||
+ | |||
+ | == 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>. | ||
+ | |||
+ | |||
+ | ==Solution 5== | ||
+ | <math>(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)</math> | ||
+ | <math>=(2(2(2(2(2(3)+1)+1)+1)+1)+1)</math> | ||
+ | <math>=(2(2(2(2(6+1)+1)+1)+1)+1)</math> | ||
+ | <math>=(2(2(2(2(7)+1)+1)+1)+1)</math> | ||
+ | <math>=(2(2(2(14+1)+1)+1)+1)</math> | ||
+ | <math>=(2(2(2(15)+1)+1)+1)</math> | ||
+ | <math>=(2(2(30+1)+1)+1)</math> | ||
+ | <math>=(2(2(31)+1)+1)</math> | ||
+ | <math>=(2(62+1)+1)</math> | ||
+ | <math>=(2(63)+1)</math> | ||
+ | <math>=(126+1)</math> | ||
+ | <math>=127 \Longrightarrow \boxed{\textbf{(C)}\ 127}</math>. | ||
+ | |||
+ | ==Solution 6 (quickest)== | ||
+ | Notice that <math>x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = x (x (x (x (x (x + 1) + 1) + 1) + 1) + 1) + 1</math>. Substituting <math>2</math> for <math>x</math>, we get <cmath>2(2(2(2(2(2+1)+1)+1)+1)+1)+1 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2 + 1 = 2^7 - 1 \Longrightarrow \boxed{\textbf{(C)}\ 127}</cmath> | ||
+ | |||
+ | ==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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:35, 4 September 2022
Contents
Problem
What is the value of ?
Solution 1
Notice this is the term in a recursive sequence, defined recursively as
Thus:
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
.
Solution 5
.
Solution 6 (quickest)
Notice that . Substituting
for
, we get
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.