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Difference between revisions of "1999 AHSME Problems/Problem 6"

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Revision as of 13:34, 5 July 2013

Problem

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 $\boxed{\text{D}}$.

See also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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
All AHSME Problems and Solutions

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