### Brain Teasers

# One Half

Arrange the digits 0, 1, 2, 2, 3, 4, 4, 5, 6, and 9, to form a single fraction that equals 1/2. (Here is one example that, alas, turns out not to be true: 32145/24690=1/2)

### Answer

One solution: 12345/24690=1/2Hide Answer Show Answer

## What Next?

View a Similar Brain Teaser...

If you become a registered user you can vote on this brain teaser, keep track of which ones you have seen, and even make your own.

### Solve a Puzzle

## Comments

Nice one Para, I liked it... my answer was one of the other possibilities... I wonder how many possibilities there are? Heh, maybe you could ask that for a bonus point

LOL!! The example turned out to be a great clue - just take 45 and 90, then work back with the numbers (3/6 = .5, 2/4 = .5, 1/2 = .5...). I wish I had the time to figure out the other solutions...

I believe that there are 36 different answers. One set is

the 6 permutations of 2, 15, and 46 in the numerator, with 4, 30, and 92 in the corresponding

positions in the denominator. Then there are 24 permutations of 1, 2, 3, and 45 in the numerator,

with 2, 4, 6, and 90 in the corresponding positions,

and finally, there are 6 more permutations of 2, 16, and 45 in the numerator

opposite 4, 32, and 90.

The key to this puzzle is noticing that the 9 must appear in the denominator (since it would have

to be opposite 8 or 9 if it were in the numerator), the corresponding position in the numerator must be a 4,

and there must be a digit capable of generating a carry in the immediately preceding position of the numerator.

Only 5 or 6 can do that. If the 6 is in the numerator, then the 3 must be in the denominator, opposite a 1,

and have the other digit capable of generating a carry in the immediately preceding position of the numerator.

That generates the first and third set of solutions. If the 6 is in the denominator, then there are no more carries

other than the one that produced the 9, so we get the second set of solutions.

the 6 permutations of 2, 15, and 46 in the numerator, with 4, 30, and 92 in the corresponding

positions in the denominator. Then there are 24 permutations of 1, 2, 3, and 45 in the numerator,

with 2, 4, 6, and 90 in the corresponding positions,

and finally, there are 6 more permutations of 2, 16, and 45 in the numerator

opposite 4, 32, and 90.

The key to this puzzle is noticing that the 9 must appear in the denominator (since it would have

to be opposite 8 or 9 if it were in the numerator), the corresponding position in the numerator must be a 4,

and there must be a digit capable of generating a carry in the immediately preceding position of the numerator.

Only 5 or 6 can do that. If the 6 is in the numerator, then the 3 must be in the denominator, opposite a 1,

and have the other digit capable of generating a carry in the immediately preceding position of the numerator.

That generates the first and third set of solutions. If the 6 is in the denominator, then there are no more carries

other than the one that produced the 9, so we get the second set of solutions.

Well, I got the given answer pretty easily. Would have been a lot harder if you hadn't just switched the 1 and 3 in the numerator to make the example! Good one though

To post a comment, please create an account and sign in.

## Follow Braingle!