Difference between revisions of "1980 AHSME Problems/Problem 15"
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== Solution == | == Solution == | ||
− | 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{(C)13}}</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>. |
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+ | 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{(C)13}}</math>. | ||
== See also == | == See also == |
Revision as of 17:33, 6 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 |
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