1989 AMC 8 Problem 1

Attempt Problem 1 of the 1989 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 1989 AMC 8 solutions, or check the answer key.

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

1.

What is the value of
(1+11+21+31+41)+(9+19+29+39+49)? \begin{aligned} &(1 + 11 + 21 + 31 + 41) \\ &\quad {}+ (9 + 19 + 29 + 39 + 49)? \end{aligned}
150150
199199
200200
249249
250250
Answer: E
Concepts:pairing and grouping
Small hint:
Look for pairs of terms that combine into round numbers
Big hint:
Pairing 11 with 49,49, 1111 with 39,39, and so on, each pair sums to 5050
Solution:
Pair the terms so each pair sums to 5050: 1+49,1 + 49, 11+39,11 + 39, 21+29,21 + 29, 31+19,31 + 19, and 41+9.41 + 9.
There are 55 such pairs, so the total is 5×50=250.5 \times 50 = 250.
Thus, the correct answer is E .
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