Difference between revisions of "2017 AMC 10A Problems/Problem 1"
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a_5 = 31*2 + 1 = 63.\\ | a_5 = 31*2 + 1 = 63.\\ | ||
a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127} | a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127} | ||
− | \end{split}</cmath> | + | \end{split}</cmath>. |
== Solution 2 == | == Solution 2 == |
Revision as of 00:34, 15 January 2018
Problem
What is the value of ?
Solution 1
Notice this is the term in a recursive sequence, defined recursively as Thus: .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.