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Fraction


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#1 rookie1ja

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Posted 30 March 2007 - 07:39 PM

Fraction - Back to the Number Puzzles
Can you arrange 9 numerals - 1, 2, 3, 4, 5, 6, 7, 8 and 9 - (using each numeral just once) above and below a division line, to create a fraction equaling to 1/3 (one third)?

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.
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#2 sn00bino

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Posted 27 April 2007 - 08:30 PM

I understand the solution. But how do you find it ? You can ' t test everything.
Sorry for my poor english
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#3 rookie1ja

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Posted 27 April 2007 - 09:23 PM

I understand the solution. But how do you find it ? You can ' t test everything.
Sorry for my poor english



A few hints:
1. first number must have 4 digits
2. second number must have 5 digits
3. second number must start with either 1 or 2
4. first number can not start with either 1 or 2
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#4 sn00bino

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Posted 28 April 2007 - 10:40 AM

Why must the second number start with 1 or 2 ?
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#5 rookie1ja

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Posted 28 April 2007 - 10:54 AM

Why must the second number start with 1 or 2 ?



Because if you multiply a 4-digit number by 3, you can not get a number which is 30000 or bigger.
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#6 sn00bino

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Posted 28 April 2007 - 11:07 AM

ok thank you
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#7 lovruch

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Posted 02 May 2007 - 09:39 PM

numerator is 4 digit, denominator is 5 digit, 5 must be in numerator and only feasible place is the thousands place, hence denominator must start with 16 or 17,
denominator can't be 16*** because the ratio has to be 1/3 and only choices for hundreds place in numerator that are 4 and 6 will violate this condition or cause repetition, hence the number is 5***/17***, now hundreds place in num can be occupied only by 8 or 9, it can't be 9 or there will be repetition, so number is 58**/17***, now just simple hit and try on the ones and tens place of numerator will result in 5832/17496
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#8 fsm

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Posted 06 May 2007 - 01:05 AM

5823/17469 also works
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#9 robxmccarthy

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Posted 25 May 2007 - 05:07 PM

This one just about made my head explode. I took the same logical work down approach, but I couldn't get the last few numbers. I kept getting one repetition.
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#10 strikefiend

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Posted 20 June 2007 - 07:28 PM

Wow, thanks for the answer. I'd been trying to use 1-9 on the top AND bottom of the problem until i gave in and had a look at the solution. Could someone check for me and see if all nine on the top AND bottom is even able to be accomplished. As far as I can tell, it's not.
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