1987 AMC 8 Problem 3

Attempt Problem 3 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).

3.

What is the value of
2(81+83+85+87+89+91+93+95+97+99)? \begin{gathered} 2(81 + 83 + 85 + 87 + 89 \\ {}+ 91 + 93 + 95 + 97 + 99)? \end{gathered}
16001600
16501650
17001700
17501750
18001800
Answer: E
Concepts:arithmetic sequencewhole number operations
Small hint:
The ten numbers 81,81, 83,83, …,\ldots, 9999 are evenly spaced
Big hint:
Their average is 81+992=90,\dfrac{81 + 99}{2} = 90, so their sum is 10×9010 \times 90
Solution:
The ten evenly spaced numbers have average 90,90, so their sum is 10×90=900.10 \times 90 = 900.
Doubling gives 2×900=1800.2 \times 900 = 1800.
Thus, the correct answer is E .
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