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May 10, 2002, 7:22:30 PM5/10/02

to

I've checked through as many groups as I have found and can't find any

answer to the following:

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

May 10, 2002, 8:00:23 PM5/10/02

to

"Barry Sanders" <bsan...@paradise.net.nz> wrote in message

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?

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.

--

remove nospam to reply

May 10, 2002, 9:42:19 PM5/10/02

to

In article <HDYC8.205$7N.2...@news02.tsnz.net>,

"Barry Sanders" <bsan...@paradise.net.nz> wrote:

"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.

May 11, 2002, 7:58:04 AM5/11/02

to

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?

multiple of 9. But can it be done for 99 instead of 100?

May 11, 2002, 8:37:37 AM5/11/02

to

"Larry Hammick" <larryh...@telus.net> wrote in message

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?

>

>

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.

May 11, 2002, 1:51:54 PM5/11/02

to

In article <WbZC8.3550$p56.1...@newsfep1-win.server.ntli.net>,

"Terry" <d...@ntlworld.com> wrote:

"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.

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