1993 AMC 8 Problem 4

Attempt Problem 4 of the 1993 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 1993 AMC 8 solutions, or check the answer key.

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

4.

1000×1993×0.1993×10=1000 \times 1993 \times 0.1993 \times 10 =
1.993×1031.993 \times 10^3
1993.19931993.1993
(199.3)2(199.3)^2
1,993,001.9931{,}993{,}001.993
(1993)2(1993)^2
Answer: E
Concepts:place valueexponent
Small hint:
Group the powers of ten: 1000×10=100001000 \times 10 = 10000
Big hint:
10000×0.1993=1993,10000 \times 0.1993 = 1993, so the product is 1993×19931993 \times 1993
Solution:
Regroup as (1000×10)(1000 \times 10) ×0.1993\times 0.1993 ×1993\times 1993 =10000= 10000 ×0.1993\times 0.1993 ×1993.\times 1993.
Since 10000×0.1993=1993,10000 \times 0.1993 = 1993, the product is 1993×1993=(1993)2.1993 \times 1993 = (1993)^2.
Thus, the correct answer is E .
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