Difference between revisions of "1980 AHSME Problems/Problem 15"
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Say that the price of the item in cents is <math>x</math> (so <math>x</math> is a positive integer as well). The sales tax would then be <math>\frac{x}{25}</math>, so <math>n=\frac{1}{100}\left( x+\frac{x}{25}\right)=\frac{26x}{2500}=\frac{13x}{1250}</math>. | Say that the price of the item in cents is <math>x</math> (so <math>x</math> is a positive integer as well). The sales tax would then be <math>\frac{x}{25}</math>, so <math>n=\frac{1}{100}\left( x+\frac{x}{25}\right)=\frac{26x}{2500}=\frac{13x}{1250}</math>. | ||
| − | Since <math>x</math> is positive integer, the smallest possible integer value for <math>n=\frac{13x}{1250}</math> occurs when <math>x=1250</math>, which gives us the answer <math>\fbox{\text{( | + | Since <math>x</math> is positive integer, the smallest possible integer value for <math>n=\frac{13x}{1250}</math> occurs when <math>x=1250</math>, which gives us the answer <math>\fbox{\text{(B)13}}</math>. |
== See also == | == See also == | ||
Latest revision as of 12:30, 9 March 2020
Problem
A store prices an item in dollars and cents so that when 4% sales tax is added, no rounding is necessary because the result is exactly 
dollars where is a positive integer. The smallest value of is
Solution
Say that the price of the item in cents is 
(so is a positive integer as well). The sales tax would then be 
, so 
.
Since is positive integer, the smallest possible integer value for 
occurs when 
, which gives us the answer 
.
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
