1998 AJHSME Problems/Problem 7

Problem 7

$100\times 19.98\times 1.998\times 1000=$

$\text{(A)}\ (1.998)^2 \qquad \text{(B)}\ (19.98)^2 \qquad \text{(C)}\ (199.8)^2 \qquad \text{(D)}\ (1998)^2 \qquad \text{(E)}\ (19980)^2$

$19.98\times100=1998$ $1.998\times1000=1998$

$1998\times1998=(1998)^2=\boxed{D}$

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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