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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)?

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Fraction - solution

5832/17496 = 1/3

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• 4 weeks later... I understand the solution. But how do you find it ? You can ' t test everything.

Sorry for my poor english

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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

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

##### Share on other sites ok thank you

##### Share on other sites 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

##### Share on other sites 5823/17469 also works

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• 3 weeks later... 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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• 4 weeks later... 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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• 2 weeks later... 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 in a random order, to create a fraction equalling 1/3 (one third)?

If you systematically choose which order to place the numbers in, then surely you cannot call it a 'random' order.

From this I deduce that Trial and error is the only way to solve this puzzle!

##### Share on other sites How bout 1^9756482 / 3.

You never said not to use any mathematical operator..

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• 2 weeks later... 3+6+9/(4+5)+18+27

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• 3 weeks later... 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.

This is how I understood the question, and I found it's certainly possible. In fact, there are 768 occurrences of this. OK, I cheated and wrote a program to figure this out

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• 3 months later... Hah. Oh.

I guess I misunderstood the problem, because this is what I came up with:

(1)(2+3)^0

(3)(4+5+6+7+8+9)^0

(which would of course reduce to

(1)(1)

(3)(1)

or

1

3

I guess that's the lazy answer.) (You could distrubute the numbers in the parentheses raised to 0 any way you want, anything raised to 0 is 1.)

edit:

How bout 1^9756482 / 3.

You never said not to use any mathematical operator..

Guess I wasn't the first person to come up with this ^^ I thought all the digits were supposed to be combined w/ operators, not as multi-digit number.

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

Umm....by placing them in any set/given order would result in them no longer being placed randomly so I don't think it can be done. Unless of course the answer for any given combination of these numbers results in 1/3 which it does not.

Just a little nitpick at the wording ;-)

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

Umm....by placing them in any set/given order would result in them no longer being placed randomly so I don't think it can be done. Unless of course the answer for any given combination of these numbers results in 1/3 which it does not.

Just a little nitpick at the wording ;-)

I have noticed that as well ... too late

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• 2 weeks later... So would this have more then one answer? or is it only one unique answer?

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• 1 month later... Also the top number has to all add up to a 3 digit number. good puzzle.

##### Share on other sites i dont think so. :huh:

##### Share on other sites 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
##### Share on other sites 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)?

Fraction - solution

5832/17496 = 1/3

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. ##### Share on other sites

• 1 month later... 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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• 2 weeks later... 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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• 2 months later... I got the same answer, though it was out of luck instead of strategy.

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.

Feels good to get this problem right when I thought it was impossible before ##### Share on other sites This topic is now closed to further replies.