1973 AMC 12 Problem 7

Attempt Problem 7 of the 1973 AMC 12 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 1973 AMC 12 solutions, or check the answer key.

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

The sum of all the integers between

50

and

350

which end in

1

5880

5539

5208

4877

4566

The integers form an arithmetic sequence from

51

through

341

Use the common difference to count the terms, then pair the first and last

The sequence is

51,61,\ldots,341.

It has

\frac{341-51}{10}+1=30

terms. Its sum is therefore

\begin{aligned} \frac{30(51+341)}2 &=15\cdot392\\ &=5880. \end{aligned}

Therefore, the correct answer is A.

Problem 7 in other years