Difference between revisions of "1993 AHSME Problems/Problem 11"

Latest revision as of 22:11, 27 May 2021

Problem

If $\log_2(\log_2(\log_2(x)))=2$ , then how many digits are in the base-ten representation for x?

$\text{(A) } 5\quad \text{(B) } 7\quad \text{(C) } 9\quad \text{(D) } 11\quad \text{(E) } 13$

Solution

Taking successive exponentials $\log_2(\log_2(x)) = 2^2 = 4$ and $\log_2(x) = 2^4=16$ and $x = 2^{16}$ . Now $2^{10} = 1024 \approx 10^3$ and $2^6 = 64$ so we can approximate $2^{16} \approx 64000$ which has 5 digits. In general, $2^n$ has approximately $n/3$ digits.

$\fbox{A}$

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 == Problem ==
-If <math>log_2(log_2(log_2(x)))=2</math>, then how many digits are in the base-ten representation for x?
+If <math>\log_2(\log_2(\log_2(x)))=2</math>, then how many digits are in the base-ten representation for x?
 <math>\text{(A) } 5\quad

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Difference between revisions of "1993 AHSME Problems/Problem 11"

Latest revision as of 22:11, 27 May 2021

Problem

Solution

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