2011 AMC 12A Problem 7

Below is the professionally curated solution for Problem 7 of the 2011 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12A solutions, or check the answer key.

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Concepts:prime factorizationdivisibilitybounding to limit cases

Difficulty rating: 1370

7.

A majority of the 3030 students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1.1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $17.71.\$17.71. What was the cost of a pencil in cents?

77

1111

1717

2323

7777

Solution:

Total cents is 1771=71123.1771 = 7 \cdot 11 \cdot 23. Writing (students)(pencils each)(cost per pencil) =1771,= 1771, the number of students is a divisor of 17711771 that is a majority of 30,30, hence more than 15.15. The only such divisor is 23.23.

Then (pencils)(cost) =77=711= 77 = 7 \cdot 11 with cost >\gt pencils >1,\gt 1, forcing 77 pencils at 1111 cents each.

Thus, the correct answer is B.

Problem 7 in Other Years