Showing the summation of numbers 
Using each of the digits 1 through 9 once, form numbers whose sum is 100. If you think it can't be done, then prove it.

My attempt:
I say it can't be done because the sum of all numbers $1-9$ is $45.$ Since this is too low (to reach $100$) you must use some $2$ figure numbers ($23, 34,$ etc.). 
But all of these give you an addition of a multiple of $9$
for instance use $'12'$ instead of $'1+2'.$ ( creates an additional 9 since you have to remove the $1$ and $2$ ).Your total is now $45 - 1 - 2 + 12 = 54$.
No matter what doubles you choose, they will always add some multiple of $9.$ 
Assume that the high part of your double is $'a'$ and the low part is $'b'$ 
$(10a + b) - a - b = 9a$
For you to succeed you need to add $55$ (to reach $100$) to your original total.
You can't do this with multiples of $9.$ You can only use $5$ once too.
I wanted to know if this proof was correct. Can someone please help me with this? Is there another way of proving this?
 A: Since the sum of the digits
is 45,
which is divisible by 9,
by the usual
casting out 9s rule,
the sum of all the numbers
has to be divisible by 9.
Since 100 is not divisible by 9,
it can never be the sum.
A: The only case we have to consider here is a series of numbers that consist of $1$ and $2$ digit numbers. First realise that any 2 digit number of the form "$xy$" (e.g. 54, 32, 15...) can be written as:
$$"xy"=10x+y.$$
Now, assume that our series consists of $n$ 2 digit numbers. Obviously $n<5$ since there can be no more than $4$ 2 digit numbers as all digits are exhausted by the attempted 5th.
Now, if there are $n$ 2 digit numbers, then this implies we can select $9-2n$ unique 1 digit numbers for our series.
We are now going to assume that there exists a series of 2 and 1 unique digit numbers that add to 100. That is
$$x_0y_0 + ... + x_ny_n+z_0+...+z_{9-2n}=100,$$
where "$x_0y_0$...." are 2 digit numbers (as you can see there are $n$ of them) and $z_0$... are 1 digit numbers. Using the representation of a 2 digit number above we can rewrite our expression as
$$(10x_0+y_0)+(10x_1+y_1)+...+(10x_n+y_n)+z_0+...+z_{9-2n}=100$$
$$10(x_0+x_1+...x_n)+(y_0+y_1+y_n)+(z_0+...+z_{9-2n})=100$$
$$10\sum_{i=0}^{n}x_i+\sum_{i=0}^{n}y_i+\sum_{i=0}^{9-2n}z_i=100$$
$$\sum_{i=0}^{n}x_i=10-\frac{\sum_{i=0}^{n}y_i+\sum_{i=0}^{9-2n}z_i}{10}.$$
Now, the LHS is clearly an integer. Furthermore, we examine
$$\sum_{i=0}^{n}y_i+\sum_{i=0}^{9-2n}z_i.$$
Notice that there are going to be $n+9-2n=9-n$ digit integers in this particular sum. Now, for the LHS to be a integer it is required that this sum results in a multiple of 10. More specifically, since $y_i$ and $z_i$ are digits, the number they must add up to is 10. The largest 1 digit sum of numbers that add $10$ is $4$, given by $1 + 2 + 3 + 4$. Therefore, 
$$9-n=4$$
$$\implies n=5.$$
This is a contradiction to the fact that $n<5$. Therefore there can exist no combination of 2 and 1 digit numbers such that the sum is 100.
