Competition problem — Number theory

Library / /38 of 520

Number theory Difficulty 4.7 AIME Find the answer

4. A merchant bought $n$ (where $n$ is a prime number) televisions for $m$ yuan (where $m$ is a natural number). Two of these televisions were sold to a charitable organization at half the cost price, and the rest were sold in the store, each with a profit of 500 yuan. As a result, the merchant made a profit of 5500 yuan. The minimum value of $n$ is:

Pick one

Solution

4. (C).

From the problem, $\frac{m}{n}+(n \cdots-2)\left(\frac{m}{n}+506\right) \cdots 5500+m$ .
Solving, we get $m=500 n(n-13)$ .
Since $m>0$ , then $n>13$ , and the smallest prime number greater than 13 is 17. Therefore, the answer is (C).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.