Difference between revisions of "2020 CIME I Problems/Problem 2"
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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>. | ||
| − | {{CIME box|year=2020|n=I| | + | {{CIME box|year=2020|n=I|num-b=1|num-a=3}} |
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
{{MAC Notice}} | {{MAC Notice}} | ||
Revision as of 20:36, 30 August 2020
Problem 2
At the local Blast Store, there are sufficiently many items with a price of 
for each nonnegative integer 
. A sales tax of 
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 
items were purchased, the total price before the sales tax is 
cents for some positive integer 
. After the sales tax of is applied, the price before tax is multiplied by 
. Thus we need 
to be an integer. This implies that ![\[\frac{4300N-43k}{40}\]](http://latex.artofproblemsolving.com/b/7/8/b78aa368928eb2fa1145a5d79861ee73a9a17a52.png)
is an integer, so 
is a multiple of 
.
This condition implies that is a multiple of because 
and are relatively prime. If is odd, the least possible value of is 
; if is even, the least possible value is . The smaller of these is obviously 
.
| 2020 CIME I (Problems • Answer Key • Resources) | ||
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