1982 AMC 12 Problem 28

Attempt Problem 28 of the 1982 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 1982 AMC 12 solutions, or check the answer key.

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

28.

A set of consecutive positive integers beginning with 11 is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is 35717.35\frac7{17}. What number was erased?
66
77
88
99
can not be determined
Answer: B
Concepts:meanarithmetic sequencedivisibility
Small hint:
If the last written number is n,n, the remaining count is n−1n-1
Big hint:
Use the bounds 1≤e≤n1\le e\le n on the erased value to narrow the possible values of nn
Solution:
The new average 60217\frac{602}{17} is below the original average n+12\frac{n+1}{2} but differs from it by less than nn−1,\frac{n}{n-1}, forcing n=69n=69 or 70.70. Since the remaining sum (n−1)(60217)(n-1)(\frac{602}{17}) must be integral, n−1n-1 is divisible by 17,17, so n=69.n=69. The erased number is 2415−68⋅60217=2415−2408=7. \begin{aligned} 2415-68\cdot\frac{602}{17} &=2415-2408\\ &=7. \end{aligned}
Therefore, the correct answer is B.
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