Difference between revisions of "2017 AMC 10A Problems/Problem 1"
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<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> | ||
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== Solution 1 == | == Solution 1 == | ||
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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>. | ||
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+ | ==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>. | ||
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+ | ==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> | ||
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+ | ==Solution 7 (Fast)== | ||
+ | Notice that there are 5 instances where the sum is multiplied by <math>2</math>. That gives us <math>2^5 = 32</math>. Separate the addition part into <math>3</math> and the <math>1</math>. Multiply the <math>3</math> by <math>32</math>, giving <math>96</math>. Then notice that the <math>1</math> is multiplied by increasing powers of two; therefore, it is equal to <math>2^5-1</math>. Then add these two parts. <math>96+31=\boxed{\textbf{(C)}\ 127}</math> | ||
==Video Solution== | ==Video Solution== | ||
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https://youtu.be/kA6W8SwjitA | https://youtu.be/kA6W8SwjitA | ||
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~savannahsolver | ~savannahsolver | ||
Latest revision as of 09:22, 1 November 2024
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
Solution 7 (Fast)
Notice that there are 5 instances where the sum is multiplied by . That gives us . Separate the addition part into and the . Multiply the by , giving . Then notice that the is multiplied by increasing powers of two; therefore, it is equal to . Then add these two parts.
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.