Difference between revisions of "2017 AMC 10A Problems/Problem 1"
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<math>=(126+1)</math> | <math>=(126+1)</math> | ||
<math>=127 \Longrightarrow \boxed{\textbf{(C)}\ 127}</math>. | <math>=127 \Longrightarrow \boxed{\textbf{(C)}\ 127}</math>. | ||
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+ | ==Solution 6== | ||
+ | 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 = 127 \Longrightarrow \boxed{\textbf{(C)}\ 127}</cmath> | ||
==Video Solution== | ==Video Solution== |
Revision as of 11:33, 4 September 2022
Contents
[hide]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
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.