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
.....................

all my +s are gone!!!

Yes, I see that. But not this time. Were you doing anything differently?

Nope

But I have had problems posting here for nearly 2 weeks.. it kept crashing (from work and home) :(

Stumped (the thread below) still gives me an error!!

CHECK:

1 + 1 = 2
1+1 = 2

I think the forum has a problem with 1_+_1 replace underscores with spaces.. as per example 1 above...

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REPOSTED:

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
.....................

I have found the answer using digits 1-9, therefore 0 comes easily here goes:
15
36
47
2
+0
__
100

This problem is in the book "More Games for the Superintelligent," by James F. Fixx.
Their version does not incorporate 0, but that dosn't matter.

Your solution has no 9.

Thanks for your work on the "proof" that it cannot be done. I copied
it and will try to understand it when I have time to really sit down
and digest it.

Meanwhile I have tried a few more times (using excel for the arithmatic) without success, of course. The thing that keeps me at
it is the fact that it was stated as a problem. This usually means
there is a solution. And if so, I enjoy working at it.

Whatever the outcome, it is fun.

Besides the fact that ehop did not include the 9 in his "solution",
I do not consider adding the 0 to be legit. I think the 0 must be part of a number - either as the second digit of a two digit number, or as part of a decimal number.

I have come as close as 99.9, but that is not hard.

Jack

>I have come as close as 99.9, but that is not hard.
Or indeed 99.999, which seems to be the closest (and is, if anything, easier).

Thankyou

sfwc
<><

> i got two answers, no need to have formula just trial and error using integrals 1 to 9 (following succession of order) to arrive to the sum of 100! in my first answer i just multiplied 2 integers and added the rest...total...100. un my 2nd anwer i multiplied 3 sets of integers and added the rest. See... its possible.. just had good luck..

Thanks for your work on the "proof" that it cannot be done.
>I copied
>it and will try to understand it when I have time to really
>sit down
>and digest it.
>
>Meanwhile I have tried a few more times (using excel for the
>arithmatic) without success, of course. The thing that
>keeps me at
>it is the fact that it was stated as a problem. This
>usually means
>there is a solution. And if so, I enjoy working at it.
>
>Whatever the outcome, it is fun.
>
>Besides the fact that ehop did not include the 9 in his
>"solution",
>I do not consider adding the 0 to be legit. I think the 0
>must be part of a number - either as the second digit of a
>two digit number, or as part of a decimal number.
>
>I have come as close as 99.9, but that is not hard.
>
>Jack

The key word here is "integers." An easy answer is:

0 + 123 + 45 - 67 + 8 - 9.

Those who think adding the zero is not legit, find a better solution :D.