Mathematical riddle?

First, mark a spot on a slope and place the balance scale there. This slope must not be steep enough for the balance scale to slide down.

Next, slide down a ball from the top of the slope towards the scale. If the scale moves after the collision, mark the spot where it moved to. If it does not move, use a steeper slope and try again.

Put back the scale to the original spot which you marked initially and repeat the experiment for each ball.

The ball that caused the scale to move to a different distance compared to the rest of the balls is the oddball.

split the balls into groups of 3. Weigh one group against another. Now either both sides are equal meaning that you know the unweighed group contains the odd ball. Or one of the weighed groups will be heavier. Actually I think you've got this riddle wrong because you normally know that the ball is heaver/lighter.

Anyway, once you've isolated the group of three with the odd ball in you can just weigh one of the 3 against another one of the 3. If they're the same then the one you didn't way is the ball of if there is a difference in the ones you weigh then you can tell which one it is.

Put 4 balls in each side of the scale & put an extra ball aside weigh it first to the right then to the left & see what will be the amount which is heavier which is lighter !!

"Think of words ending in -GRY. Angry and hungry are two of them. There are only three words in the English language. What is the third word? The word is something that everyone uses every day. If you have listened carefully, I have already told you what it is."

Erm, about ure riddle: let's say it's heavier

first weigh three balls in one pan against three balls in the other. If the pans balance, the heavier ball must be among the last three. If the pans don't balance, the heavier ball must be among the three in the heavier pan. So, in the first weighing, itself, I would have narrowed the location of the golden ball to a group of three.

"I would then take two of these three balls and place one in each pan. If the pans balance, the heavier ball would be the last one.

If the pans don't balance, the heavier ball would be the one in the heavier pan. Thus, the heavier ball can be located in two weighings only."

First weigh 6 balls, three balls per side.

If they are balanced, take 2 of the remaining balls and weigh those, if they are balanced it's the remaning ball.

I can't think of a way to find the remaining ball unless I know if it is either heavier or lighter.

weigh eight balls, 4 on each side, then take the heavier/lighter side and weigh those 4, two on each side, (if the are both the same, then the one that you didn't weigh is the bad one) take the two that are on the heavier/lighter side and then you have to guess

that reminds me a of a mathematical joke..

Two octogenarians decided to take separate vacations. They kept in contact during their vacations, each insisting they were having a better time than the other. It seems each had taken up with a 40 year old lover.

Finally the old wife explained why she was indeed having a better time than her 80-something husband:

"It's simple mathematics dear.. really...40 goes into 80 more times than 80 goes into 40.

It all sounds a load of balls to me. That's a "don't know " from me

FIRST chance-- weigh 9... SECOND-- invert to 6 and weigh again.....just a guess....thx

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