1987 AMC 8 Problem 10

Attempt Problem 10 of the 1987 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 1987 AMC 8 solutions, or check the answer key.

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

10.

What is the value of
4(299)+3(299)+2(299)+298?4(299) + 3(299) + 2(299) + 298?
28892889
29892989
29912991
29992999
30093009
Answer: B
Concepts:distributive property
Small hint:
Combine the first three terms: 4(299)+3(299)+2(299)4(299) + 3(299) + 2(299) =9×299= 9 \times 299
Big hint:
Compute 9×299,9 \times 299, then add 298298
Solution:
The first three terms combine to (4+3+2)(299)(4 + 3 + 2)(299) =9×299=2691.= 9 \times 299 = 2691.
Adding the last term gives 2691+298=2989.2691 + 298 = 2989.
Thus, the correct answer is B .
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