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Old 2007-08-07, 16:18   #1
mfgoode
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Question Student error.


A student wrote out this equation

123456789 = 100.

Can you insert plus and minus signs in between digits of the number on LHS. to make it right ? You could use even two digits together. Its not necessary to only take single digits and the plus and minus signs need not be equal in number.

Try it at the most for 7 signs.

Can you do it with only 3 plus or minus signs ?

Mally

Last fiddled with by mfgoode on 2007-08-07 at 16:23
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Old 2007-08-07, 18:08   #2
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123-45-67+89 = 100


What if we allow the other operations and parentheses?

For example, 1+2+3+4+5+6+7+8*9 = 100.

Are there other such representations?
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Old 2007-08-07, 18:20   #3
axn
 
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Code:
123+45-67+8-9=100
123+4-5+67-89=100
123-45-67+89=100
123-4-5-6-7+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
1+23-4+56+7+8+9=100
1+23-4+5+6+78-9=100
1+2+34-5+67-8+9=100
1+2+3-4+5+6+78+9=100
Attached Files
File Type: txt puzzle.txt (40.2 KB, 114 views)

Last fiddled with by axn on 2007-08-07 at 18:30
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Old 2007-08-08, 11:39   #4
mfgoode
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Smile Other operators.

Quote:
Originally Posted by davar55 View Post

123-45-67+89 = 100


What if we allow the other operations and parentheses?

For example, 1+2+3+4+5+6+7+8*9 = 100.

Are there other such representations?


In this problem I only gave the simplest of cases and the problem can be complicated enough, by using fractions and allowing zero as the tenth digit,and not allowing other operations like multiplications.

The example you have given does not qualify for using *. But you may lump two digits together and maintain the rule to only the use of a max. of 7 signs. HINT: make use of the minus sign too.

You are correct in the 3 sign rule. Excellent work!

If you can get the 7 sign one then these two solutions ( 7signs and 3 signs) are the only few of the very few solutions possible for the 9 digits. I have got only two of the 7 sign ones at hand. There may be more in this computer era as these problems were based on pre-computer days.

We will then extend the problem to fractions and ten digits.

Best of luck!

Mally

Last fiddled with by mfgoode on 2007-08-08 at 11:42
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Old 2007-08-08, 12:12   #5
mfgoode
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Question The works !

Quote:
Originally Posted by axn1 View Post
Code:
123+45-67+8-9=100
123+4-5+67-89=100
123-45-67+89=100
123-4-5-6-7+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
1+23-4+56+7+8+9=100
1+23-4+5+6+78-9=100
1+2+34-5+67-8+9=100
1+2+3-4+5+6+78+9=100
Well axn1 you amaze me. I did not check your attached file when I replied to Davar 55. Is that the lot your comp. can cough up ? Or are there more?

One observation with strained eye I have made. You have also given Davar's three sign one. In the good old days this was considered the only combination we could get. It will be worthwhile to find another.

I mean 123 - 45 -67 + 89 = 100.

Best of luck!

Mally
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Old 2007-08-08, 13:22   #6
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Some other ( may be all solutions), which include multiplication and division operations also, described at http://hi.baidu.com/%C9%BD%C9%BD%BA%...61d9583b0.html

Last fiddled with by VolMike on 2007-08-08 at 13:24
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Old 2007-08-09, 17:27   #7
axn
 
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Quote:
Originally Posted by mfgoode View Post
Well axn1 you amaze me. I did not check your attached file when I replied to Davar 55. Is that the lot your comp. can cough up ? Or are there more?

One observation with strained eye I have made. You have also given Davar's three sign one. In the good old days this was considered the only combination we could get. It will be worthwhile to find another.

I mean 123 - 45 -67 + 89 = 100.
I have attached the pascal program I've used to brute-force this one. Except for the fact that I haven't allowed for a leading minus sign (which can be easily done, btw), the program does a thorough search. The result file I had attached earlier was the output of the program sorted and truncated upto 200. There were a number of other totals, but that didn't interest me too much. In total, there are exactly 3^8 = 6561 ways of putting + & - in between the 9 digits.

So, to answer your question, what I had posted earlier was the exhaustive list of solutions for 100 (again, the case of a leading minus sign is not accounted for). So there is only one solution with three signs.
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File Type: zip puzzle1.zip (489 Bytes, 87 views)
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Old 2007-08-10, 06:08   #8
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I suppose it wouldn't be too much more difficult using a brute-force method, but what about solutions using add-subtract-multiply-divide where you evaluate like in RPN [ie, 1+23-45*678-9 would be (((1+23)-45)*768)-9]?
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Old 2007-08-10, 08:35   #9
mfgoode
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Thumbs up Pascal program.

Quote:
Originally Posted by axn1 View Post
I have attached the pascal program I've used to brute-force this one. Except for the fact that I haven't allowed for a leading minus sign (which can be easily done, btw), the program does a thorough search.

So, to answer your question, what I had posted earlier was the exhaustive list of solutions for 100 (again, the case of a leading minus sign is not accounted for). So there is only one solution with three signs.


Thank you axn1 once again.

I am not into programming at all and perhaps it will be a diversion to me, and too late a stage in my life, to take it up now. However people like you, who have mastered both ends of the spectrum of modern Maths, are what is required today.

If you have at least one sign + in your solutions along with the minus terms, can't you put it first so that there is no need specifying this in your programme as a minus to be leading ? I'm sure this can be done.

So Old Kordemsky who wrote "The Moscow Puzzles' back in the old days was right after all that the 3 sign solution was, and hearing from you probably, the
only solution. It will be great if you coud step up your search for another 3 sign solution It will make history!

Mally
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Old 2007-08-10, 08:44   #10
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Post Gallop ahead!

Quote:
Originally Posted by Kevin View Post
I suppose it wouldn't be too much more difficult using a brute-force method, but what about solutions using add-subtract-multiply-divide where you evaluate like in RPN [ie, 1+23-45*678-9 would be (((1+23)-45)*768)-9]?


Thank you Kevin for extending the problem to ' divide multipy' solutions.

We can do that but lets finish off with plus and minus solutions as I have got a few related problems dealing in two signs Plus and minus, which we can tackle. This is a result of skimming thru all the puzzles books I have and these are considerable in number ( I dont even know how many). And Im presenting only number theory puzzles for our members as a collection of them for ready reference. I will be back to present some more problems

Mally
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Old 2007-08-10, 16:53   #11
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Question Fractions.



Well I am back again.

1) Using the digits 0 to 9, only once each, find fractions with plus or minus signs to total to 100.

Eg: 1 3/6 + 98 27/54 + 0 = 100

2)What is the unique whole number whose square and cube between them use up each of the digits from 0 to 9 ?

3) Compose two fractions using digits from 0 to 9 whose sum shall be equal to unity ?

4) Given 6 different digits forming a number beginning with 28.

By transfering 28 to exteme rght hand side we get the new number thus formed which is exactly double the original number Find the number.

Mally
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