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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?

Michael Li
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2 Answers2

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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.

Benjamin Wang
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  • Can you give me a mathematically written proof so that I can see this process? Thanks. – Michael Li Jul 27 '20 at 02:26
  • Sure. Edited to give an example. – Benjamin Wang Jul 27 '20 at 02:46
  • wow, thank you! – Michael Li Jul 27 '20 at 02:56
  • This is the official solution: Suppose k items were purchased, so that the price before tax is 100N −k cents for some integer N. After a tax of 7.5%, i.e. a multiplier of 43/40 is applied, we need (43/40)(100N − k) to be an integer. It is equivalent for 100N − k to be a multiple of 40. Thus k is a multiple of 20, but k = 20 is achievable by taking N a sufficiently large odd number. Can you explain how they obtained 100N-k? – Michael Li Jul 27 '20 at 04:38
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    Sure. In 100N-k, it comes from the fact that buying each additional item decrements the last two digits. For example, if k = 3, your spending is guaranteed to be [some integer amount of dollars] + 97 cents. When stated in cents only, this is 100N-k for some integer N. (Also, please consider accepting my answer if it helped. Thanks :) ) – Benjamin Wang Jul 27 '20 at 06:10
  • Why is this true? – Michael Li Jul 27 '20 at 06:20
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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.

Michael Li
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  • Yea this is correct. Because you referred to an "answer key" (as opposed to "solution manual", I thought this question was a "insert-a-number" instead of "full solution" question. Anyway, Your last paragraph should just say "arbitrary" instead of "random" $k$. Good work :) I'm sure there's a shorter solution though. – Benjamin Wang Jul 28 '20 at 00:36
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    O, it wasn't a solution problem, I just gave one to check the validity of my logic. Thanks! Ya, my sol. was bit of an overkill. – Michael Li Jul 28 '20 at 00:38