Old-fashioned numbers game puzzle

[Johnny Canuck]

http://wiki.apterous.org/Episode_1504

It's now fairly well known that in the old days, CECIL could generate 100 as a target. In the episode linked to above, 100 also showed up as one of the large numbers, resulting in the (literally) easiest numbers game ever. To avoid the obvious 100 = 100 solution, Richard told the contestants to find a method that didn't involve actually using the 100.

Your challenge is to create a numbers game in which 100 = 100 is the only possible solution, or failing that, one where you have to use the 100 to achieve the target.

And no, there's no prize.

[Steve Balog]

Seems fairly simple: 100 1 1 2 2 3 --> 100 -- I believe 27 is the largest you can get with the smalls here. Might be a bit harder to try and make one with a two-large selection, with one 100, where 100 = 100 is the only solution. I'll think about that one. 4 large is impossible for obvious reasons, and 3 large would require the 50 as one of the other two.

EDIT: Also the second numbers solution is almost as easy hahaha, looks like CECIL didn't much believe in the contestants' math skills that day

[Steve Balog]

looking at three large:

The other larges, as said before, have to be 25 50 or 50 75.

1) For the smalls, there CANNOT be anyway, whatsoever, for the smalls to make a 2, because you can then go (75*2-50) or (25*2+50). For 25, 4 and 6 are also off-limits for similar reasons, and 4 is also off-limits for 75 50, but 6 is ok. I'll try 50 75 solutions.

100 75 50 9 9 10 -> 100 works. without the 100, only 99 is possible.

now two large:

Again, I like 75 as the large here, as 50 makes 2 unusable, and 25 4. You can't cheese it with two tens for obvious reasons, and we need to make sure the sum can't be 25. Unfortunately, the solution this time is to cheese it the other way: 100 75 2 2 1 1 -> is not only not solvable without the 100, you can't even score -- 111 is the best you can do with the other 5.

EDIT: Also, I've determined that 4 large is impossible to only have 100 = 100, 75-25+50 and 25+75 = 100 as solutions:

if 2 (50*2), 3 (3*25+75-50), 4 (25*4), 5 (25*5-75+50), 6 (25*6-50), 7 (25*7-75), or 9 (25*9-50-75) appear, then it is solvable. This leaves only 1, 8, and 10 for possible smalls.

1 + 1 = 2
1 + 8 = 9
10 - 1 = 9
10 - 8 = 2
10*10 = 100
8 8 is the hardest, but there is, for instance, 75*8/(8 - (50/25)) = 100, which is non-trivial and uses the 8's.