29 replies to this topic

[stormpool]

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!

[mannu]

How bout 1^9756482 / 3.

You never said not to use any mathematical operator..

[ahem cuba]

3+6+9/(4+5)+18+27

[awesselink]

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

[icthy]

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.

[cbrowninnc]

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 ;-)

[rookie1ja]

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

[spiralangell]

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

[HelllxWeaver]

Also the top number has to all add up to a 3 digit number. good puzzle.

[void28]

i dont think so.