1999 AHSME Problems/Problem 6

Revision as of 23:40, 26 May 2019 by Bigpoppainyomomma (talk | contribs) (Solution)

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

Solution 2

We proceed with eeny meeny miny moe, if lucky we obtain $\boxed{\text{D}}$.

See also

1999 AHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png