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Today’s Problem: Sealed Solution

Hello 4KM and 4KJ,

I think you already know I love a maths challenge. Seeing your post for your special visitors, I saw the included challenge and thought I would share my solutions. For others, here was the challenge for the students and special visitors.

This problem is from the NRICH website.

There are 10 cards each bearing the digits 0 to 9. Two are placed in each of five envelopes. The sum of the two cards in each envelope must match the number on the envelope. The challenge was to find what two cards could be in the envelope numbered 8. The students and visitors were informed there was more than one solution. See the below graphic...

 

How did I find my solutions?

1. The challenge stated the two cards in each envelope are added to reach the number on the envelope (the sum of the two cards).

2. Being ten cards each bearing a digit from 0 to 9, I assumed each digit could only be used once.

3. Next I looked at the target, Envelope 8. Using the cards, what 2 card combinations could give the sum of 8? Here they are..

8 + 0 = 8

7 + 1 = 8

6 + 2 = 8

5 + 3 = 8

4. Now I drew up a grid to test each of the sums to see if all four could be correct. I found three worked but one failed. Here are the three successful answers...

3 + 4 = 7    8 + 0 = 8   6 + 7 = 13   9 + 5 = 14   1 + 2 = 3

2 + 5 = 7    7 + 1 = 8   9 + 4 = 13   8 + 6 = 14   3 + 0 = 3

7 + 0 = 7    5 + 3 = 8   9 + 4 = 13   8 + 6 = 14   1 + 2 = 3

Therefore, Envelope 8 could only contain 0 and 8, 1 and 7, or 3 and 5.

 

Why did 2 and 6 fail?

Envelope 14 can only contain 8 and 6 or 9 and 5 to give the sum of 14. As  2 and 6 (6 + 2 = 8) have been used, 8 and 6 aren't possible but what about 9 + 5 =14?

Envelope 13 can only contain 9 and 4, 8 and 5 or 7 and 6 if the sum is to be 13. If we have already used 2 and 6 (2 + 6 = 8) and 5 and 9 (5 + 9 = 14), we can't use any solution for Envelope 13. Therefore 6 + 2 = 8  is not a solution.

How did I do?