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2011 AMC 12A Problem 7

Problem 7 of 25EasierAlgebraNumber Theory

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?

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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.

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Concepts: prime factorization · divisibility · bounding to limit cases

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.