This is in response to Jack...

I think the situation you propose, uniquely using the digits 0 to 9, combined into decimal or integers to form the addends which sum to 100 is impossible.

I think this can be developed into a properproof by contradiction, of sorts. I am at work right now, so can't do it properly, so here os where I would start.

Proof (of sorts…)

We require a series of addends composed from the digits 0 to 9 that give us the sum 100.

Let's first consider the simplest case, where all the addends are one-digit numbers, eg 'units only', namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Let K be the sum of addends.

Sum of addends: K = 0 1 2 3 4 5 6 7 8 9 = 45

Let u be the sum of the 'units' in addends.

Thus we have, for this case, K = u = 45. We require K = 100, thus this arrangement is incorrect.

Now, let's look at combining the digits into two-digit numbers, with t as the sum of the 'tens' in the addends. EG Let's look at: 0, 1, 32, 4, 5, 6, 7, 8, 9.

K = 0 1 32 4 5 6 7 8 9 = 72....(I)
u = 0 1 2 4 5 6 7 8 9 = 42.....(II)
t = 3..........................................(III)

Adding equation (II) and (III) we obtain: u t = 45 (you can varify this for all combinations of two-digit numbers - in the first case t=0).

u t = 45
u = 45 - t

We require: K = 100 = 10t u

K = 10t (45 - t)

100 = 10t (45 - t)

55 = 9t

t = 55 / 9

Since t, by definition is the sum of positive naturals – it is not possible, as t must be a positive natural too.

QED?

Extend this for decimals (1 dp).

t = sum of tens digits
u = sum of units digits
v = sum of tenths digits

t u v = 45.......................(IV)
10t u v/10 = 100.................(V)

(V) * 10 => 100t 10u v = 1000....(VI)

(VI) – (IV) => 99t 9u = 955

99t 9v = 955

9(11t v) = 955

11t v = 955/9

Since t and u, by definition are the sums of positive naturals – it is not possible, as t and u cannot be summed to give a non positive natural.

t = sum of tens digits
u = sum of units digits
v = sum of tenths digits
w = sum of hundredths digits
.....................