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Difference between revisions of "1980 AHSME Problems/Problem 15"

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== Problem ==
 
== Problem ==
  
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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 <math>n</math> dollars where <math>n</math> is a positive integer. The smallest value of <math>n</math> is
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A store prices an item in dollars and cents so that when <math>4\%</math> sales tax is added, no rounding is necessary because the result is exactly <math>n</math> dollars where <math>n</math> is a positive integer. The smallest value of <math>n</math> is
  
 
<math>\text{(A)} \ 1 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 25 \qquad \text{(D)} \ 26 \qquad \text{(E)} \ 100</math>
 
<math>\text{(A)} \ 1 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 25 \qquad \text{(D)} \ 26 \qquad \text{(E)} \ 100</math>
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== Solution ==
 
== Solution ==
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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>.  
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Let <math>x</math> be the price of the item in cents. The sales tax is <math>.04x = \frac{x}{25}</math> cents, 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{(B)13}}</math>.
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Since <math>x</math> is positive integer, the smallest possible integer value of <math>n</math> occurs when <math>x=1250</math>, and in this case <math>n</math> is <math>\fbox{\text{(\textbf{B}) 13}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:19, 15 August 2025

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 $n$ dollars where $n$ is a positive integer. The smallest value of $n$ is

$\text{(A)} \ 1 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 25 \qquad \text{(D)} \ 26 \qquad \text{(E)} \ 100$


Solution

Let $x$ be the price of the item in cents. The sales tax is $.04x = \frac{x}{25}$ cents, so $n=\frac{1}{100}\left( x+\frac{x}{25}\right)=\frac{26x}{2500}=\frac{13x}{1250}$.

Since $x$ is positive integer, the smallest possible integer value of $n$ occurs when $x=1250$, and in this case $n$ is $\fbox{\text{(\textbf{B}) 13}}$.

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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