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# Fraction

29 replies to this topic

### #21 kiger

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Posted 27 February 2008 - 05:30 AM

Can you use all 9 numerals - 1, 2, 3, 4, 5, 6, 7, 8 a 9 - above and below a fraction symbol, to create a fraction equalling 1/3 (one third)?

oh oh, i got it. errr,
yes!
word it better maybe?
kiger

Edited by kiger, 27 February 2008 - 05:31 AM.

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### #22 Bummie

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Posted 03 March 2008 - 05:55 AM

Fraction - Back to the Number Puzzles
Can you use all 9 numerals - 1, 2, 3, 4, 5, 6, 7, 8 a 9 - above and below a fraction symbol, to create a fraction equalling 1/3 (one third)?

Spoiler for Solution

I took the same approach you did to solving: numerator = 4 digits, denominator = 5 digits. Working backward: 6/18 = 1/3, 24/72 = 1/3 and 3/9 = 1/3; therefore 6243/18729 = 1/3. Took less than a minute to solve.
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### #23 Chrijenn

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Posted 17 April 2008 - 06:31 PM

I took the same approach you did to solving: numerator = 4 digits, denominator = 5 digits. Working backward: 6/18 = 1/3, 24/72 = 1/3 and 3/9 = 1/3; therefore 6243/18729 = 1/3. Took less than a minute to solve.

Although your solution works out to 1/3, your answer is incorrect as you used the numeral 2 twice and did not use the numeral 5. I beleive you were supposed to all numbers only once.
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### #24 dbm1

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Posted 30 April 2008 - 11:41 PM

the question doesn't state that the numerals must be used once and once only.

123456789/123456789(3) is probably the most obvious answer, but obviously it requires sentience, and anything sentient is not random (as someone has already mentioned). whoever writes these questions should have some understanding of semantics (the most important part of language logic).
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### #25 mayiko

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Posted 06 July 2008 - 02:26 AM

I got the same answer, though it was out of luck instead of strategy.

[spoiler='"']

I figured the numerator would have 4 digits and the denominator 5 digits (I don't think there's another setup to distribute the 9 digits). Also since 3x3=9 and 2x3=6 why don't we just use 32 as the last two digits of the numerator to get 96 as the last two digits of the denominator, so there wouldn't be so many numbers left to work with.

So we have 1, 4, 5, 7, 8 left to work with to place in the first two digits of the numerator and the first three digits of the denominator. The first digit of the denominator has to be 1 (because multiplying 4,5,7,8 by 3 gives you a first digit of 1 or 2 and 2 was already used) Then you see that by multiplying 8 by 3, you get a 4 as the second digit (so I placed 8 as the 2nd digit of the numerator and 4 as the third digit, and the 5 and 7, luckily for me, happened to fit in the rest. So I got 5832/17496.

[/spoiler]

Feels good to get this problem right when I thought it was impossible before
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### #26 cmabb21

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Posted 08 July 2008 - 07:57 PM

So would this have more then one answer? or is it only one unique answer?

well no, there are at least two answers
5823/17469
and
5832/17496
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### #27 Gazza7

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Posted 13 October 2008 - 03:34 AM

hey guys this is my first post.
You know you've made the problem harder than it really is
Try this solution4It's basic but logical

((5-4)(2-1)(7-6)(9-8))/3
=(1x1x1x1)/3
=1/3
Simple
but effective
comment if u wish
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### #28 njkellykid

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Posted 23 October 2008 - 05:36 PM

Fraction - Back to the Number Puzzles
Can you use all 9 numerals (Edit: each numeral just once) - 1, 2, 3, 4, 5, 6, 7, 8 a 9 - above and below a fraction symbol, to create a fraction equalling 1/3 (one third)?

does it have to be only numerals? can you use parentheses and and functions to, like adding or subtracting? If so, what about this

Spoiler for My solution

Spoiler for Solution

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### #29 parikshit

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Posted 20 July 2010 - 01:52 PM

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### #30 Dej Mar

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Posted 25 July 2010 - 01:31 AM

If one tries to solve this problem without electronic assistance, one can reduce their trial and error attempts with the following information.

As the ratio of the numerator to the denominator is 1/3, our denominator needs to have the same number or one more digit than the denominator, and because there are nine digits this means the numerator will have four digits and denominator will have five. The max carry for 3n, where n is a digit 1 to 9, is 2, therefore the denominator must begin with a 1 or 2. And, by the same token, the smallest number the numerator can begin with is 4. (It can not begin with 3 as 3×3 equals 9, the largest decimal digit, and adding only 1 or 2 will bring the sum to only 10 or 11. There can neither be a 0 (zero) or two 1's in a zeroless pandigital common fraction of 9 digits, therefore the numerator must begin with a 4 or greater.)

For the ratio of 1/3, the denominator also needs to be a multiple of 3, therefore its digital root must be either 3, 6 or 9, and the sum of the digits are modulo 0.

Though this puzzle has already been solved, I shall still post the solution I discovered:
Spoiler for There are two solutions:

Edited by Dej Mar, 25 July 2010 - 01:33 AM.

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