1998 AMC 8 Problem 7

Attempt Problem 7 of the 1998 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

100×19.98×1.998×1000=100\times 19.98\times 1.998\times 1000=
(1.998)2(1.998)^2
(19.98)2(19.98)^2
(199.8)2(199.8)^2
(1998)2(1998)^2
(19980)2(19980)^2
Answer: D
Concepts:decimalexponent
Small hint:
Pair the factors with powers of 10.10.
Big hint:
Group the product as (100⋅19.98)(1000⋅1.998).(100\cdot 19.98)(1000\cdot 1.998).
Solution:
Group the first two factors and the last two factors:
(100⋅19.98)(1.998⋅1000)=1998⋅1998=(1998)2. \begin{aligned} &(100\cdot 19.98)(1.998\cdot 1000) \\ &= 1998\cdot1998 \\ &= (1998)^2. \end{aligned}
Thus, the correct answer is D .
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