Sum of digits 1 - 9
Barry Sanders
answer to the following:
Using ALL the digits 1 - 9 each ONCE ONLY, combine them such that the
numbers made SUM to 100.
There are many combinations that use a COMBINATION of operators, but none
that I can find using only addition.
Is there a solution?
Is it unique?
Is there a program or method for finding such a sum?
Any help would be appreciated.
Barry Sanders
Terry
news:HDYC8.205$7N.2...@news02.tsnz.net...
> I've checked through as many groups as I have found and can't find any
> answer to the following:
>
> Using ALL the digits 1 - 9 each ONCE ONLY, combine them such that the
> numbers made SUM to 100.
>
> There are many combinations that use a COMBINATION of operators, but none
> that I can find using only addition.
>
> Is there a solution?
No!
Consider the following..
The sum of all numbers 1-9 is 45
Since this is too low you must use some 2 figure numbers (23, 34 etc)
but *all* of these give you an addition of a multiple of 9
i.e use 12 instead of 1+2 ( = 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 your original total. You can't do
this with multiples of 9.
T.
--
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The World Wide Wade
"Barry Sanders" <bsan...@paradise.net.nz> wrote:
> I've checked through as many groups as I have found and can't find any
> answer to the following:
>
> Using ALL the digits 1 - 9 each ONCE ONLY, combine them such that the
> numbers made SUM to 100.
>
> There are many combinations that use a COMBINATION of operators, but none
> that I can find using only addition.
>
> Is there a solution?
> Is it unique?
> Is there a program or method for finding such a sum?
I assume you are talking about sums like 32 + 56 + 1 + 9 + 47 + 8? If so, I
don't see a solution. Think about it this way: 1 + 2 + ... + 9 = 45. When
we try to form numbers that sum to 100, some of the digits will be in the
ten's place, the others will be in the one's place. Can 9 appear in the
ten's place? Clearly not, because then the sum is at least 90 + (45-9). You
can similarly rule out an 8 or 7 in the 10's place. How about a 6? Putting
all other digits in the 1's spot gives 99, too small, and moving another
digit over to the 10's spot gives too big a sum. Proceding in this way, you
can rule out the other possibilities.
--WWW.
Larry Hammick
multiple of 9. But can it be done for 99 instead of 100?
Terry
news:gH7D8.4352$27.1...@news2.telusplanet.net...
> Terry is right, the sum will always be a multiple of 9, and 100 is not a
> multiple of 9. But can it be done for 99 instead of 100?
>
>
In many ways :)
1+2+3+4+7+8+9+65 = 99
1+2+4+5+7+8+9+63 = 99
12+3+4+56+7+8+9 = 99
etc.
The World Wide Wade
"Terry" <d...@ntlworld.com> wrote:
> No!
>
> Consider the following..
> The sum of all numbers 1-9 is 45
> Since this is too low you must use some 2 figure numbers (23, 34 etc)
> but *all* of these give you an addition of a multiple of 9
> i.e use 12 instead of 1+2 ( = 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.
That the sum is always a multiple of 9 is a nice observation. Another way
to write your proof: Let d1,d2, ... dk be the digits you put in the 10's
place. Let S = d1 + d2 + ... dk. Then the sum of the numbers so constructed
is
10*S + 45 - S = 1*S + 0 - S = S - S = 0 (mod 9).
So the sum always = 0 (mod 9), which is the same as saying it's always a
multiple of 9.
--WWW.