Here's my best shot
If the 10 ntegers have an average of 6.....then the total sum of the integers must be 10 * 6 = 60 because 60 / 10 = 6
Therefore......we want to make the sum of the "non-one" digits as small as possible and the number of ones as large as possible
Possible Number of 1's Sum of the other digits
1 59
2 58
3 57
4 56
5 55
6 54
7 53
8 52
9 51
10 50
The last is clealrly not possible ( all ones)
The next-to-the last is also not possible......the largest sum that could be created with 9 ones is 9 + 14 = 23
The third-to-the last is also not possible ...the largest possible sum = 8 ones + 2 * 14 = 36
The fourth-to-the-last is also not possible....the largest sum = 7 ones + 3 * 14 = 49
Six ones is possible because 14 + 14 + 13 + 13 = 54
So....the max number of 1s = 6
