Can anyone work this teaser out?

[detective]

Assuming hymn cards carry a number on one side only, what is the minimum number that need to be stocked in order to cater for every possible hymn number from 1 to 999 inclusive, when up to four hymns may be sung at any given service?

[crofter]

I make it 89 cards if only one side is used. Over four hymns you would need 9 cards with "5" on to cover hymn 555 and three others with two 5's. Same goes for 1's, 2's, ... , 9's making 9x9 =81 cards. The 0's are special since they are not generally placed at the front of any hymn number, so only 8 of these are needed! This will cover hymns 100, 200, 300, 400 or similar. This brings the total to 89.

BUT if the verger is shrewd he/she can make an upsidedown 9 look like a 6 and do away with the nine cards with 6 on.

Hence I offer the answer 80 as the minimum number required.

[Pantheon]

ah yes, BUT

the verger has a favourite song / the organist only knows one music score .....

so , they sing four 'repeat' hymns ....

111+111+111+111 = 12 number ones for a start!!

[Icntoan]

I don't think you can do away with all the 6s. What if you had 999,996,699 and 696. You would need twelve cards to do this, so I think you need to add another three to the total.

I agree with Crofter on the rest so I think the answer is 83.

[crofter]

Icntoan is quite right, I overlooked the possibility that all four hymns were combinations of only 6's and 9's.

Well spotted!