Focus on the first 5 of the 6 slots to be filled with digits. If one considers two plates with identical digits on each of these slots then clearly they cannot differ in more than one place: the 6-th. So producing plates which differ at least in one place within the first 5 slots is a must. One the other hand the maximum number of such plates is 10⁵=100,000 since we consider 10 digits and 5 slots.

Assume we have produced those 100,000 plates leaving the 6-th slot empty. If we can invent an algorithm to fill out the 6-th slot in such a manner that each 2 of the 100,000 plates will differ in 2 places or more, then the answer is 100,000.

As it often happens in such problems an algorithm exists. For instance, for each plate ad up the digits on the first 5 slots and write on the 6-th slot the last digit of the number you obtained this way.

Lets show that all 100,000 plates differ in 2 places or more. Pick two of them at random. If the first 5 slots differ in two or more places, there is nothing to prove. But, what if they differ in only one place? Then the difference of the two numbers one obtains by adding the digits on the first 5 slots for each plate is the difference of the distinct digits and hence is positive and at most 9. Therefore the numbers obtained by addition of the first 5 digits are numbers with distinct last digits and hence the two plates differ on slot number 6 too.

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