CIME Money problem Question: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.
After solving this problem numerous times, I keep getting that the answer is 40, but the answer key says its 20. What am I doing wrong?
 A: Yep, it's 20.
The 7.5% means you multiply by $1+ \frac{3}{40}$.
You are forgetting that multiples of $40$ can also end in $80$.
In particular, you should try to hit a total of {an even number}.80.
So you just need $20$ (well chosen) items to get the last two digits down to $80$.
The easiest way is 19 copies of 0.99, and then one item worth 1.99.
A: Actually I think I have an answer myself!
Sol.
First off, we can convert the money into cents, namely $(100n+99)$ for all non-negative $n$'s. After the tax, we get $\frac{43}{40} \cdot (100n+99).$
Let $M = m_1 + m_2 + \ldots + m_n$ where $m_{i}$ $i\in\{1, 2, 3, \ldots,n\}$ can be any non-negative integer.
We then get \begin{align*}
(99)(\frac{43}{40})m_1 + (100+99)(\frac{43}{40})m_2 + \ldots + (100n+99)(\frac{43}{40})m_n \\
\rightarrow \frac{43}{40}(99 m_1 + 100 m_2 + 99 m_2 + 100n \cdot m_n +99m_n) \\
\end{align*}
Factoring the $99$'s and $100$'s we get $\frac{43}{40}(99M +100(M-m_1)) \implies \frac{43}{40}(199M -100m_1)$
This means that $199M-100m_1 \equiv 39M-20m_1 \equiv 0 \pmod{40} \iff 39M-20m_1 \equiv 39M \equiv 0 \pmod{4}, 39M-20m_1 \equiv 39M \equiv 0 \pmod{10}$
Since 39 doesn't contribute anything to M, we know that $M = \text{lcm}(4,10)k$ for some random $k.$ Therefore the least $M$ that satisfies the conditions is $\boxed{20}.$
A bit overkill I would say lol.
