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Difference between revisions of "2020 CIME I Problems/Problem 2"
 
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If <math>k</math> items were purchased, the total price before the sales tax is <math>100N-k</math> cents for some positive integer <math>N</math>. If the sales tax of <math>7.5\%</math> is applied, the price before tax is multiplied by <math>\frac{43}{40}</math>. Thus we need <math>\frac{43}{40}(100N-k)</math> to be an integer. This implies that <cmath>\frac{4300N-43k}{40}</cmath> is an integer, so <math>4300N-43k</math> is a multiple of <math>40</math>.  
 
If <math>k</math> items were purchased, the total price before the sales tax is <math>100N-k</math> cents for some positive integer <math>N</math>. If the sales tax of <math>7.5\%</math> is applied, the price before tax is multiplied by <math>\frac{43}{40}</math>. Thus we need <math>\frac{43}{40}(100N-k)</math> to be an integer. This implies that <cmath>\frac{4300N-43k}{40}</cmath> is an integer, so <math>4300N-43k</math> is a multiple of <math>40</math>.  
 
This condition implies that <math>100N-k</math> is a multiple of <math>40</math> because <math>43</math> and <math>40</math> are relatively prime. If <math>N</math> is odd, the least possible value of <math>k</math> is <math>20</math>; if <math>k</math> is even, the least possible value is <math>40</math>. The smaller of these is obviously <math>\boxed{20}</math>.
 
This condition implies that <math>100N-k</math> is a multiple of <math>40</math> because <math>43</math> and <math>40</math> are relatively prime. If <math>N</math> is odd, the least possible value of <math>k</math> is <math>20</math>; if <math>k</math> is even, the least possible value is <math>40</math>. The smaller of these is obviously <math>\boxed{20}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=gdOnSyFewDU
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~Shreyas S
  
 
==See also==
 
==See also==

Latest revision as of 23:03, 4 September 2020

Problem 2

At the local Blast Store, there are sufficiently many items with a price of $$n.99$ for each nonnegative integer $n$. A sales tax of $7.5\%$ is applied on all items. If the total cost of a purchase, after tax, is an integer number of cents, find the minimum possible number of items in the purchase.

Solution

If $k$ items were purchased, the total price before the sales tax is $100N-k$ cents for some positive integer $N$. If the sales tax of $7.5\%$ is applied, the price before tax is multiplied by $\frac{43}{40}$. Thus we need $\frac{43}{40}(100N-k)$ to be an integer. This implies that \[\frac{4300N-43k}{40}\] is an integer, so $4300N-43k$ is a multiple of $40$. This condition implies that $100N-k$ is a multiple of $40$ because $43$ and $40$ are relatively prime. If $N$ is odd, the least possible value of $k$ is $20$; if $k$ is even, the least possible value is $40$. The smaller of these is obviously $\boxed{20}$.

Video Solution

https://www.youtube.com/watch?v=gdOnSyFewDU ~Shreyas S

See also

2020 CIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All CIME Problems and Solutions

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