Onehundred from ten

4 + 20 + 7 + 8 + 9 + 56 - (1 + 3) = 100
1 + 60 + 2 + 5 + 8 + 37 - (4 + 9) = 100

The parenthetical expression must sum to 4 or 13.

If that is so, there are only 3 more possible solutions?

3 + 12 + 4 + 5 + 9 + 80 - (6 + 7 ) = 100
_ + _ _ + _ + _ + _ + _ _ - (4 + 0) = 100
_ + _ _ + _ + _ + _ + _ _ - (8 + 5) = 100

You can always swap tens digits in the two 2-digit numbers.

12, 80 -> 10, 82

and units digits in a 1-digit number and a 2-digit number.

3, 12 -> 2, 13

I don't know if any other rearrangements of the non-parenthetical numbers will work.

O right. So 10 solutions per parenthetical expression, of which there are 5.


How do you know that the parenthetical expression must sum to 4 or 13?

Whim

Background information: The digital root of N is (N mod 9). The digital root of N is equal to the digital root of the sum of the digits of N. The digital root of M+N is the digital root of M + the digital root of N. The digital root of M*N is the digital root of M * the digital root of N. If you haven't heard of digital roots, you've probably heard of casting out nines.

The digits 0-9 sum to 45 (45 mod 9 = 0), so the digital root of any additive combination/concatenation of them must be zero.

Since we have to subtract two of the digits, the result will have a digital root of 0 - (sum of the two digits because we didn't add them) - (sum of the two digits because we're subtracting them). Therefore, the digital root of the result will be 0 - 2*(sum of the two digits). The digital root of 100 is 1.

Call the subtracted digits x and y. We have:
1 = -2*(x+y) mod 9
-1 = 2*(x+y) mod 9 (divide by -1)
8 = 2*(x+y) mod 9 (8 = -1 mod 9)
4 = (x+y) mod 9 (divide by 2)
Therefore,
x+y = 4
x+y = 13
x+y = 22
etc.

However, since x+y <= 17, we only have to deal with the first two possibilities.