Infinate
7-1
8-2
9-3
etc
An infinite number.
There are 36 (6 times 6) outcomes. These comprise 11 sums or differences, 18 products etc.
4-6=2x9=18 divided by 2=15
You can get 2, 4, 6, 8, 10, and 12, six even sums all together.
8 quarters equals 2
Well, 1 is equal to 1. 9 is equal to 9. So any number from 0 to 9. I do not think this is what is meant by the question. If you mean any number with more than one digit. There is no such number. Because 11 sums to 2. 19 sums to 10. 21 sums to 3. 29 sums to 11. 111 sums 3. 119 sums to 11. The sums do not grow as fast the increase of digits.
23
The positive integer factors of 3246 are: 1, 2, 3, 6, 541, 1082, 1623, 3246 Limiting the combinations of sums merely to multiplication sums using positive integer factors, the following sums can be made: 1 x 3246 = 3246 2 x 1623 = 3246 3 x 1082 = 3246 6 x 541 = 3246
9...7 + 6 + 14 = 27 divided by 3 (number of sums) = average I would hope this is a joke
Turnaround facts help you solve sums by encouraging you to start with the greater number first and add the lesser. ie. 5 + 1 = 6 is easier to solve than 1 + 5 = 6
Add all the sums, then divide by the number of sums. (ie. the average.)
All the possible sums are: 6 7 8 9 11 12 13 16 17 21