Difference between revisions of "1971 AHSME Problems/Problem 14"
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<cmath>2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)</cmath> | <cmath>2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)</cmath> | ||
− | We only care about two terms: <math>2^ | + | We only care about two terms: <math>2^6+1</math> and <math>2^6-1</math>. These simplify to <math>65</math> and <math>63</math>. |
Thus, our answer is <math>\boxed{\textbf{(C) }63,65}.</math> | Thus, our answer is <math>\boxed{\textbf{(C) }63,65}.</math> |
Revision as of 12:53, 1 August 2024
Problem
The number is exactly divisible by two numbers between
and
. These numbers are
Solution
Factor with difference of squares.
We only care about two terms: and
. These simplify to
and
.
Thus, our answer is
-edited by coolmath34
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.