1999 AHSME Problems/Problem 6: Difference between revisions

Latest revision as of 22:59, 16 March 2020

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

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1999 AHSME Problems/Problem 6: Difference between revisions

Latest revision as of 22:59, 16 March 2020

Problem

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