112k views
2 votes
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.

User Jan Hudec
by
7.9k points

1 Answer

2 votes
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. :)
User Dhanaji Yadav
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories