Ans 21
Ans 19
The numbers can only be made of powers of 5. and the sum of their powers will be 13.
Total non-negative solutions of three powers adding to 13 is 15C2 = 105
But they have repetitions...
0,0,13
1,1,11
2,2,9
3,3,7
4,4,5
5,5,3
6,6,1
Each of these 7 sets will have 3 permutations... so they take up 7*3=21 out of 105 solutions.
The remaining 84 will have all three elements different and will have 6 permutations each so they will give 84/6 = 14 unique sets.
So answer is 14+7 = 21 such sets of numbers.
For 5^12,
total solutions is 14C2 = 91
4,4,4 will occur once so 1 solution
0,0,12
1,1,10
2,2,8
3,3,6
5,5,2
6,6,0
These six sets will occur thrice each so 6*3= 18 solutions
Remaining 72 solutions will be made of all different powers that gives 72/6 = 12 unique sets.
So total of 1+6+12 = 19 solutions. |
