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Difference between revisions of "1971 AHSME Problems/Problem 1"

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== Solution ==
 
== Solution ==
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<math>N=2^{12}\cdot5^8=2^4\cdot2^8\cdot5^8=2^4\cdot(2\cdot5)^8=2^4\cdot10^8=16\cdot100000000</math>, which has <math>2+8=10</math> digits, so the answer is <math>(\textbf{D})</math>.
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<math>N=2^{12}\cdot5^8=2^4\cdot2^8\cdot5^8=2^4\cdot(2\cdot5)^8=2^4\cdot10^8=16\cdot100000000</math>, which has <math>2+8=10</math> digits, so the answer is <math>(\textbf{B})</math>.

Revision as of 01:49, 13 July 2018

Problem 1

The number of digits in the number $N=2^{12}\times 5^8$ is

$\textbf{(A) }9\qquad \textbf{(B) }10\qquad \textbf{(C) }11\qquad \textbf{(D) }12\qquad \textbf{(E) }20$

Solution

$N=2^{12}\cdot5^8=2^4\cdot2^8\cdot5^8=2^4\cdot(2\cdot5)^8=2^4\cdot10^8=16\cdot100000000$, which has $2+8=10$ digits, so the answer is $(\textbf{B})$.

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