If you do not need to use all the digits then there are lots of simple solutions: 92+8, or example.
If you can use operations other than addition, again, there are numerous soultions:
0 + 1 + 23 - 4 + 56 + 7 + 8 + 9 = 100.
or
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 +8*9 = 100.
If the only operation permitted is addition, and all ten digits are to be used then it is relatively easy to show that it cannot be done. But first you need to know about digital roots. The digital root of a number is obtained by adding up all its digits. If the answer is bigger than 10 you repeat until you get to a number in the range 0-9.
For example, the digital root of (456789) is 4+5+6+7+8+9 = 39, which in turn gives 3+9 =12 which gives 1+2=3.
The numbers 0-9 add to 45 which has a digital root of 4+5 = 9. So ANY sum which contains all the digits 0-9, whether in the units place or tens place (or hundreds or millions for that matter), will have a digital root of 9. But 100 has a digital root of 1. And that mismatch is enough.
A more rigorous proof would require some knowledge of modulo 9 arithmetic but that is going too far.
If
For example, by adding 1 + 1 + 1 ... (a total of 100 times).
The numbers are 55.5 and 44.5
60
200+100
They are: 50+50 =100 or 10 times 10 = 100
1+23-4+56+7+8+9+100
Adding: 100+44 Multiplying: 12*12
For example, by adding 1 + 1 + 1 ... (a total of 100 times).
Well, isn't that just a happy little question! When you add 100 plus 100 together, you get 200. It's like adding two friendly little numbers that come together to create something beautiful. Just imagine those numbers dancing together on your canvas, creating a perfect sum of 200.
4. 4x25=100.
123 + 45 -67 + 8 - 9 = 100!
The sum of the numbers 1 to 100 is 5050.
The numbers are 55.5 and 44.5
60
There are no two prime numbers in which a product of 100 is possible.
10 + (9 x 8) + 7 + 6 + 5 + 4 - 3 - 2 + 1 = 100
2,17,23,29,29