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| raed |
 Posted: Wed May 07, 2008 7:21 am Post subject: Try that ! |
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Below is an equation that isn't correct yet.
By adding a number of plus signs and minus signs between the numbers on the left side (without changes the order of the numbers), the equation can be made correct.
123456789 = 100
The Question: How many different ways are there to make the equation correct?
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Raed Ahmed Shalaby
Dont let MEDIA affect you , You have a brain !
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| JaneBennet |
| Posted: Wed May 07, 2008 8:41 am Post subject: |
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Do the signs have to be strictly between the numbers – or can you put a minus sign in front of the 1? For example:
−1+2−3+4+5+6+78+9 = 100
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Last edited by JaneBennet on Wed May 07, 2008 9:27 am; edited 2 times in total |
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| raed |
| Posted: Wed May 07, 2008 8:45 am Post subject: |
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| JaneBennet wrote: |
can you put a minus sign in front of the 1?
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Yes , u can
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Raed Ahmed Shalaby
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| JaneBennet |
| Posted: Wed May 07, 2008 11:01 am Post subject: |
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These are what I’ve found so far:
- −1+2−3+4+5+6+78+9 = 100
- 1+2+34−5+67−8+9 = 100
- 1+2+3−4+5+6+78+9 = 100
- 1+23−4+5+6+78−9 = 100
- 1+23−4+56+7+8+9 = 100
- 12−3−4+5−6+7+89 = 100
- 12+3−4+5+67+8+9 = 100
- 12+3+4+5−6−7+89 = 100
- 123−4−5−6−7+8−9 = 100
- 123+4−5+67−89 = 100
- 123+45−67+8−9 = 100
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| raed |
| Posted: Wed May 07, 2008 12:27 pm Post subject: |
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Nice , How long did it take u to do that ?? 
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Raed Ahmed Shalaby
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| JaneBennet |
| Posted: Wed May 07, 2008 12:58 pm Post subject: |
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About an hour using Excel. 
I’m still not sure if I’ve found all the solutions though. Maybe Serpico can tell us how to solve the problem using only logical deductive reasoning. 
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| bit4bit |
| Posted: Wed May 07, 2008 12:59 pm Post subject: |
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I've had a crack at this problem, but haven't solved it. What I thought was that first of all, find out how many possible combinations there are of the left hand side, whether they =100 or not, and then somehow (But I don't know how) find the number that does =100 using that. This is what i've done. I'm probably not on the right lines, but I'll share it anyway
So each digit can be represented in a kind of binary, where it can have two states, +ve or -ve. For example:
1-2+3-4+5-6+7-8+9 can be written as:
(+1)+(-2)+(+3)+(-4)+(+5)+(-6)+(+7)+(- +(+9)
so you add up all the digits, which can be either positive or negative. So considering individual digits only, there is 29 combinations.
Then you can have 'blocks' of digits, for example:
1+2+3+4+5+6+78+9
So you need to add the number of combinations for different sized blocks. But you can also have more than one block in the left hand side. For the left hand side only having one block in it (ranging from block size 1 to block size 9), I got:
29+8*28+7*27+6*26+5*25+4*24+3*23+2*22+2
Then the next step would be to add the combinations for having more than one block in the left hand side, and having combinations of different sized blocks in the left hand side, but I haven't done that yet.
I'm trying to use algebra to make things less tedious. I can be sure that what i'm trying to calculate is gonna be a very large number, but I'm not sure if it helps things at all. I'm finding this to be quite an interesting problem anyway.
What do you think? Is there some hint as to what kind of things to use to solve it? I mean what area of maths deals with this? Is it a graph theory problem?
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Last edited by bit4bit on Wed May 07, 2008 1:04 pm; edited 1 time in total |
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| bit4bit |
| Posted: Wed May 07, 2008 1:02 pm Post subject: |
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| JaneBennet wrote: |
| About an hour using Excel. |
Thats cheating 
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| raed |
| Posted: Wed May 07, 2008 1:07 pm Post subject: |
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| bit4bit wrote: |
| What do you think? Is there some hint as to what kind of things to use to solve it? I mean what area of maths deals with this? Is it a graph theory problem? |
U are talking as if I Know the answer ! 
Really i have no idea about the answer of that problem
But that is nice , Let's think together
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Raed Ahmed Shalaby
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| bit4bit |
| Posted: Wed May 07, 2008 1:12 pm Post subject: |
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Oh I thought it was one of those 'quiz' questions.
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| JaneBennet |
| Posted: Wed May 07, 2008 1:17 pm Post subject: |
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| bit4bit wrote: |
| JaneBennet wrote: |
| About an hour using Excel. |
Thats cheating |
I know. 
As for the calculation of how many ways you can insert pluses and minuses between the digits, I think of it this way. There are 8 spaces between the 9 digits, and each of the spaces can have 3 states: +, −, or nothing. When there is nothing between two digits, they combine into a “block”. This gives 38 possible ways of putting + and − between some or all of the spaces between the digits. Moreover, Raed has said that you can add a minus sign to the 1 at the beginning. Hence the total number of ways of putting + and − according to the rules of the problem is 2 × 38 = 13 122. 
| raed wrote: |
Really i have no idea about the answer of that problem
But that is nice , Let's think together |
Well, we know so far that there are at least 11 solutions to the problem. 
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“A problem worthy of attack
Proves its worth by fighting back.” – Piet Hein
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| bit4bit |
| Posted: Wed May 07, 2008 2:37 pm Post subject: |
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| JaneBennet wrote: |
| bit4bit wrote: |
| JaneBennet wrote: |
| About an hour using Excel. |
Thats cheating |
I know.
As for the calculation of how many ways you can insert pluses and minuses between the digits, I think of it this way. There are 8 spaces between the 9 digits, and each of the spaces can have 3 states: +, −, or nothing. When there is nothing between two digits, they combine into a “block”. This gives 38 possible ways of putting + and − between some or all of the spaces between the digits. Moreover, Raed has said that you can add a minus sign to the 1 at the beginning. Hence the total number of ways of putting + and − according to the rules of the problem is 2 × 38 = 13 122. |
Oh yeh, Thanks! That's so much simpler. The number I would have gotten from what I wrote is 4098, so it wasn't as big as I thought it looked. Maybe the method I was using would have arrived at the same answer as yours. I'm pretty certain it's a correct method, but obviously it's way too tedious.
I can't think of anything for the next step to start eliminating possibilities....without having to try all 13,122 of them. What a brain teaser!
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| river_rat |
| Posted: Fri May 09, 2008 8:28 am Post subject: |
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I get 12 (and according to matlab that is all there is - if you want the code shout)
My List
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89
-1+2-3+4+5+6+78+9
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| raed |
| Posted: Fri May 09, 2008 8:41 am Post subject: |
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How could you solve it using Matlab ?
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Raed Ahmed Shalaby
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| JaneBennet |
| Posted: Fri May 09, 2008 8:46 am Post subject: |
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| river_rat wrote: |
| I get 12 (and according to matlab that is all there is - if you want the code shout) |
Ooh, I missed one!
| Quote: |
My List
1+2+3-4+5+6+78+9
1+2+34-5+67-8+9
1+23-4+5+6+78-9
1+23-4+56+7+8+9
12+3+4+5-6-7+89
12+3-4+5+67+8+9
12-3-4+5-6+7+89
123+4-5+67-89
123+45-67+8-9
123-4-5-6-7+8-9
123-45-67+89
-1+2-3+4+5+6+78+9 |
That was the one I missed. Thanks.
_________________
“A problem worthy of attack
Proves its worth by fighting back.” – Piet Hein
Why can’t a bull see red – literally can’t? Did You Know? |
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