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1999 AHSME Problems/Problem 6: Difference between revisions

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<math> \textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10</math>
<math> \textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10</math>
==Solution==
<math>2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}</math>, a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be <math>2+5=7</math>, thus making the answer <math>\boxed{D}</math>.

Revision as of 19:09, 2 June 2011

What is the sum of the digits of the decimal form of the product $2^{1999}\cdot 5^{2001}$?

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$

Solution

$2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}$, a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be $2+5=7$, thus making the answer $\boxed{D}$.

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