I was initially puzzled, because 6 items of price 1-6 gives 6^6 possible combinations, or 46656 combinations. If each of these costs 1 cent more than the one below it, then the highest priced combination would cost $466.56, and not $1,176.48. It took me some time to figure out if this can be done. It can, almost.

The solution in the end is simple: take base 6, instead of base 7. Then, if all costs 1 cents, you pay 1+6+.36+216+1296+7776 =9331 cents, or in base 6

111111 cents.

Then you go up to 111112, 11113, etc, up to 111120 if the 1st item costs 6 cents.

If the 2nd item costs 2 cents, we would read

111121, so this wouldn’t overlap with this previous possibility.

The way to read the answer is a bit more complicated, but hints at why this works.

Take the price, subtract 111111 base 6. Then, as in the previous answer, read the price off digit by digit, but then add one to each digit to get the real price.

So, 111120-111111=000005 (base 6), and therefore the prices are 1,1,1,1,1,6

and 111121-111111=000010 (base 6), so the prices are 1,1,1,1,2,1.

This has you pay a maximum of $559.86

Still quite a lot, but less.

I think this is the optimal answer, though theoretically $466.56 is the lowest possible.

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Hi! I was pleasantly surprised to her my name on the podcast :)!

I think that the answer is universal. The only two things you need to know is the highest possible cost item and the total number of items. The highest cost plus 1 is the base system used (in the problem 6+1 = 7) and the number of items determines the order in which this is done (like the number of columns before). I do not even think that the prices need to be ordered. I mean, it could be 10 items from 1-100 cents or 100 items from 1-10 cents. My convictions come from the fact that it is not using a pre-determined base (like the decimal system of 10) but adjusts to the problem. Also, each value is determined independently of each other, there is no influence from other answers.

I could be wrong though …

Also, amusingly, I am like Juho as I am currently visiting a university in Finland. (Jyväskylä … have fun pronouncing that!) ]]>

The technique is totally correct, but does it require that no item cost more than your n? Our problem was set up a little differently– we know the cost is as much as 10 cents, and the technique we have will work up to 10 items. In other words, n has to be the max of the costs and the number of possible different items.

Unless I’m missing something obvious of course!

Thanks for a great email,

Chaim

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