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
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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