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I came across this sign in a shop and thought it could make a nice puzzle. So you can buy items and get discounts depending on how many items you got. You can combine discounts and use as many as you need. I want to buy $n$ items. Is there a value for $n$, such that I need to use all three types of discounts to get the best savings?

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No, there isn't.

As you can always favourably replace a 2 and 4 by two 3s the best bargain cannot contain 4s and 2s at the same time.

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  • $\begingroup$ Correct and well done! $\endgroup$ – Dmitry Kamenetsky May 7 at 14:31
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    $\begingroup$ Moreover, you never want to use more than two 4's as three 4's are worse than four 3's. This type of things happen very often in real life. Sometimes I have to argue a lot at checkout. $\endgroup$ – WhatsUp May 8 at 2:23
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    $\begingroup$ Once a shoe shop had an event saying that buying two items and the third (cheapest) was free. I bought six items, and at checkout they made the two cheapest items free... I argued that it should be the third and the sixth being free, but they couldn't understand. In the end I separated them into two groups and checked-out twice... $\endgroup$ – WhatsUp May 8 at 2:27
  • $\begingroup$ My father once saw a sale of \$1 items at 4 for \$5. At checkout, he requested that the clerk ring them up at the regular price. $\endgroup$ – RobPratt May 10 at 13:39

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