2011 AMC 12A Problems/Problem 7: Difference between revisions
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== Solution == | == Solution == | ||
The total cost of the pencils can be found by <math>(\text{students}*\text{pencils purchased by each}*\text{price of each pencil})</math> | The total cost of the pencils can be found by <math>(\text{students}*\text{pencils purchased by each}*\text{price of each pencil})</math>. | ||
Since <math>1771</math> is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: <math>7, 11, 23</math>. Since neither <math>(C)</math> nor <math>(E)</math> are any of these factors, they can be eliminated immediately, leaving <math>(A)</math>, <math>(B)</math>, and <math>(D)</math>. | |||
Beginning with <math>(A) 7</math>, we see that the number of pencils purchased by each student must be either <math>11</math> or <math>23</math>. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this. | Beginning with <math>(A) 7</math>, we see that the number of pencils purchased by each student must be either <math>11</math> or <math>23</math>. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this. | ||
Revision as of 23:44, 26 January 2019
Problem
A majority of the 
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 
. 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 
. What was the cost of a pencil in cents?
Solution
The total cost of the pencils can be found by 
.
Since 
is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: 
. Since neither 
nor 
are any of these factors, they can be eliminated immediately, leaving 
, 
, and 
.
Beginning with 
, we see that the number of pencils purchased by each student must be either 
or 
. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.
Continuing with 
, we can conclude that the only case that fulfills the restrictions are that there are students who each purchased 
such pencils, so the answer is 
. We can apply the same logic to as we applied to , if one wants to make doubly sure.
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
