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put this solution on YOUR website!Suppose we use two three-digit numbers.
To get 5 as a units digit, we need either 4+1, 3+2. Since 9+6 = 5 and 8+7 = 5 (mod 10) we cannot use 8 and 9 for the units digit (since we would need a 6 or 7).
Therefore we have either {4,1} or {3,2} as the units digits. Note that the tens digits have to be 8 and 9 since putting them in the hundreds place would create numbers too large. For the hundreds digits, use the set that wasn't used for the units digits (e.g. if 4,1 are units digits, use 3,2). Due to carrying, the hundreds digit will always be 6.
This now becomes a combinatorics problem where we assign digits to each number. We can evaluate the number of ways using combinatorics, but we have to account for overcounting and commutativity (e.g. 281+394 = 394+281), so I'll just list them out:
183 + 492
182 + 493
193 + 482
192 + 483
281 + 394
284 + 391
291 + 384
294 + 381
This leaves 8 ways.
Now, suppose we use a three-digit number, a two-digit number, and a one-digit number. This is obviously impossible, since the largest value we can obtain is 498 + 32 + 1 < 675. Additionally, no other combination of digits (e.g. four-digit plus two-digit) can work, so we have 8 ways.
