1967 AMC 12 Problem 30

Attempt Problem 30 of the 1967 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 1967 AMC 12 solutions, or check the answer key.

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

30.

A dealer bought nn radios for dd dollars, d,d, a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of $8\$8 on each radio sold. If the overall profit was $72,\$72, then the least possible value of nn for the given information is:
1818
1616
1515
1212
1111
Answer: D
Concepts:moneylinear equationdivisibility
Small hint:
Each radio costs dn\frac{d}{n} dollars
Big hint:
Write the total intake from n−2n-2 profitable sales and two half-cost sales
Solution:
The n−2n-2 regular sales bring (n−2)(dn+8),(n-2)(\frac{d}{n}+8), while the two bazaar radios bring dn\frac{d}{n} together. Since the total intake is d+72,d+72, (n−2)(dn+8)+dn=d+72. \begin{aligned} &(n-2)\left(\frac dn+8\right)+\frac dn\\ &\qquad=d+72. \end{aligned} This reduces to d=8n(n−11).d=8n(n-11). Positivity requires n>11,n\gt11, and n=12n=12 gives the positive integer d=96.d=96.
Therefore, the correct answer is D.
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