Question

Using the digits 1 to 9, at most one time each, fill in the blanks to make two different pairs of two-digit numbers that have a sum of 71.

Answer 1

Do you mean no repeating numbers within the two sets? Because if so, I don't think it's possible.

I started by trying to figure out what the numbers in the tenths place should be. I used subtraction: 71 - ___ = ____. If you try it out, you can't subtract anything with a 6, 7, 8, or 9 in the tenths place because it will leave you with 11 (a repeating digit number), 10 (has a 0), or less (1-digit numbers). Also, a 3 cannot go into the tenths place because when you do 71 - 3_ , your answer will always begin with a 3 (problem because the 3 repeats), or it will contain a 0. 

Therefore, the numbers left for the tenths place are: 1, 2, 4, and 5. 1 and 5 pair up, leaving 4 and 2.
71 - 5_ = 1_   and 71 - 4_ = 2_.

Then, I tried to figure out what numbers go in the ones place. That lead me to realize they act in pairs. The pairs possible in the ones place are 2 and 9, 3 and 8, 4 and 7, 5 and 6. These numbers always go together to result with the final "1" in the "71". Using this information, I looked at the numbers I already used: 1, 2, 4, and 5. Now, looking at the pairs, I eliminated the pairs containing a number already used. This leaves me with only one pair: 3 and 8. Obviously, you need two more pairs to solve the problem, which leads me to my point of saying: This problem is impossible to solve.

I really hope someone can prove me wrong! But this is the solution I have reached for now. :)