View source for 1991 AJHSME Problems/Problem 3

==Problem==

Two hundred thousand times two hundred thousand equals

<math>\text{(A)}\ \text{four hundred thousand} \qquad \text{(B)}\ \text{four million} \qquad \text{(C)}\ \text{forty thousand} \qquad \text{(D)}\ \text{four hundred million} \qquad \text{(E)}\ \text{forty billion}</math>

==Solution==

Writing the numbers in scientific notation, we have
<cmath>\begin{align*}
(2\times 10^5)\times (2\times 10^5) &= 4\times 10^{10} \\
&= 40\times 10^9 \\
&= \text{forty billion} \rightarrow \boxed{\text{E}}.
\end{align*}</cmath>

==See Also==

{{AJHSME box|year=1991|num-b=2|num-a=4}}
[[Category:Introductory Algebra Problems]]
{{MAA Notice}}