2020 CIME I Problems/Problem 2

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.


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 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All CIME Problems and Solutions

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