How Can I Make A Total Of 37 Using Only 10 Numbers From 8 1's, 9 3's, 8 5's, and 8 7's ?
I initially did some math and found out there isn't any solution with exactly 10 numbers:
suppose there are A 1s, B 3s, C 5s, and D 7s
A+B+C+D=10A+3B+5C+7D=37
so:
A=10-B-C-D
10-B-C-D+3B+5C+7D=37
2B+4C+6D=27
2(B+2C+3D)=27
left side is even, right side is odd
No solution to the problem.
Then I wrote a little program to find solution assumin A+B+C+D<10
It's not possible with exactly 10 numbers, but possible with less (22 solutions):
0 one 0 three 6 five 1 seven
0 one 1 three 4 five 2 seven
0 one 2 three 2 five 3 seven
0 one 3 three 0 five 4 seven
0 one 4 three 5 five 0 seven
0 one 5 three 3 five 1 seven
0 one 6 three 1 five 2 seven
1 one 0 three 3 five 3 seven
1 one 1 three 1 five 4 seven
1 one 2 three 6 five 0 seven
1 one 3 three 4 five 1 seven
1 one 4 three 2 five 2 seven
1 one 5 three 0 five 3 seven
2 one 0 three 0 five 5 seven
2 one 0 three 7 five 0 seven
2 one 1 three 5 five 1 seven
2 one 2 three 3 five 2 seven
2 one 3 three 1 five 3 seven
3 one 0 three 4 five 2 seven
3 one 1 three 2 five 3 seven
3 one 2 three 0 five 4 seven
4 one 0 three 1 five 4 seven
I like the way you have done that with a little computer program. I assume that is how you did it.
I initially did some math and found out there isn't any solution with exactly 10 numbers:
suppose there are A 1s, B 3s, C 5s, and D 7s
A+B+C+D=10A+3B+5C+7D=37
so:
A=10-B-C-D
10-B-C-D+3B+5C+7D=37
2B+4C+6D=27
2(B+2C+3D)=27
left side is even, right side is odd
No solution to the problem.
Then I wrote a little program to find solution assumin A+B+C+D<10