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

Revision as of 19:16, 2 June 2011

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
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

@@ Line 10: / Line 10: @@
 == See also ==
-{{AHSME box|year=1999|numb-b=5|num-a=7}}
+{{AHSME box|year=1999|num-b=5|num-a=7}}
 [[Category:Introductory Algebra Problems]]

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

Revision as of 19:16, 2 June 2011

Problem

Solution

See also