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1991 AJHSME Problems/Problem 3

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Problem

Two hundred thousand times two hundred thousand equals

\text{(A)}\ \text{four hundred thousand} \qquad \text{(B)}\ \text{four million} \qquad \text{(C)}\ \text{forty thousand} \qqu...

Solution

Writing the numbers in scientific notation, we have \begin{align*}(2\times 10^5)\times (2\times 10^5) &= 4\times 10^{10} \\&= 40\times 10^9 \\&= \text{forty billion}...

See Also

1991 AJHSME (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
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Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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