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 ?

Guest Oct 28, 2014

#4**+10 **

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

Guest Oct 28, 2014

#2**+5 **

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

Guest Oct 28, 2014

#3**0 **

I like the way you have done that with a little computer program. I assume that is how you did it.

Melody
Oct 28, 2014

#4**+10 **

Best Answer

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

Guest Oct 28, 2014