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>.
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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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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 13:30, 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 $n$ 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

Say that the price of the item in cents is $x$ (so $x$ is a positive integer as well). The sales tax would then be $\frac{x}{25}$, 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 for $n=\frac{13x}{1250}$ occurs when $x=1250$, which gives us the answer $\fbox{\text{(B)13}}$.

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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