1991 AMC 8 Problem 4

Attempt Problem 4 of the 1991 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 1991 AMC 8 solutions, or check the answer key.

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

4.

If 991+993+995991 + 993 + 995 +997+999+ 997 + 999 =5000−N,= 5000 - N, then N=N =
55
1010
1515
2020
2525
Answer: E
Concepts:arithmetic sequence
Small hint:
Each of the five numbers is just a little less than 10001000
Big hint:
Write each as 1000−k;1000 - k; the sum becomes 5000−(9+7+5+3+1)5000 - (9+7+5+3+1)
Solution:
Since 991+993+995+997+999=(1000−9)+(1000−7)+(1000−5)+(1000−3)+(1000−1)=5000−(9+7+5+3+1)=5000−25, \begin{aligned} &991+993+995 \\ &\quad {}+997+999 \\ &= (1000-9)+(1000-7) \\ &\quad {}+(1000-5) \\ &\quad {}+(1000-3)+(1000-1) \\ &= 5000 - (9+7+5+3+1) \\ &= 5000 - 25, \end{aligned} we get N=25.N = 25.
Thus, the correct answer is E .
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