Thread: Riddle Me This, SRP

Since I trust Gugi implicitly, yes, I am satisfied with his answer.

Next riddle please!!

James.

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Originally Posted by xman

@$$#ø|€

calligraphy?

I shall warn you that this one may be a bit challenging - it took me about 20min the first time I heard it, but I like a good challenge. If it fails I'll come up with another one


you have 13 balls and one of them is different weigh than the rest.
with only 3 measurements using simple scales, i.e. equal weight or not between groups of balls, find the odd ball.

Originally Posted by Jimbo

Since I trust Gugi implicitly, yes, I am satisfied with his answer.

And I guess this confession also makes your trust explicit from now on

Originally Posted by gugi

I shall warn you that this one may be a bit challenging - it took me about 20min the first time I heard it, but I like a good challenge. If it fails I'll come up with another one


you have 13 balls and one of them is different weigh than the rest.
with only 3 measurements using simple scales, i.e. equal weight or not between groups of balls, find the odd ball.

Are you told which side is heaviest, or just given binary info i.e. same vs not the same?

Also, can you "cheat" by observing what happens as you remove one ball from each side after each weighing?

Originally Posted by Jimbo

OK. Here's one that should get you going..

If n is an integer greater than two prove, without recourse to the Taniyama-Shimura Conjecture, that the equation a^n + b^n = c^n has no solutions for non-zero integers a, b, and c.

Note: you may write your solutions in the margin if you wish. If the margin should prove too narrow to contain your proof, you may leave your solution tantalisingly unwritten for several centuries.

James.

OK I can prove it. Next question?

Lets call the 1-13. Ignore the thirteenth one for now.

First Usage:

Weigh 4 weights on each side (remember which ones and the slope):
eg. 1-2-3-4 vs 5-6-7-8
Unbalance --> < A > (answer in one of these 8 weights)
Balance --> < B > (answer in other 4 weights)

Second Usage
< A >
Of the 8 weights you weighed before, take 3 off the right side, and
grab 2 from the left side and put it them on the right. And then
one weight you didn't weigh before on the left.
eg. 1-2-x vs 3-4-5
(where x=9, 10, 11 or 12)
Slope of Balance Unchanged -->
Slope of Balance Changed -->
Balance -->

< B >
Leave 3 weights on one side, and take 3 weights you didn't weigh
before and put them one the other side.
eg. 1-2-3 vs 9-10-11
Right Heavier -->
Left Heavier -->
Balance --> ANSWER = Weight never weighed

Third Usage

Since the slope is unchanged, the two you switched from left to
right before and are not the answer. The answer is one of the other
three weights, the other 9 are "wrong". Now, move the "answer"
weight from the right and put it on the left, and take the other
"answer" weight (on the left) out. Now put two of the other "wrong"
weights on the right side.
eg. 1-5 vs x-x
(where x=3, 4, 6, 7, 8, 9, 10, 11 or 12)
Slope of Balance Changed --> ANSWER = Weight on right in 2nd
usage
Slope of Balance Unchanged --> ANSWER = Weight on left in 2nd
usage
Balance --> ANSWER = Weight left out


Since the slope has changed, the one of two weights you switched
from left to right before are the answer. Move one of the two back
to the left side.
eg. 3 vs 4
Slope of Balance Changed --> ANSWER = Left weight
Slope of Balance Unchanged --> ANSWER = Right Weight


Since the scale is balanced, the answer is one of the 3 weights you
had taken out. Put one on each side of the scale and leave the
other out.
eg. 6 vs 7
Unbalance --> If slope in First Usage said that right side was
heavier, ANSWER = Heavier weight
--> If slope in First Usage said that right side was
lighter, ANSWER = Lighter weight
Balance --> ANSWER = Weight left out


The three on the right are heavier, so put one on each side of the
scale, and leave one out.
eg. 9 vs 10
Unbalance --> ANSWER = Heavier side
Balance --> ANSWER = Weight left out


The three on the right are lighter, so put one on each side of the
scale, and leave one out.
eg. 9 vs 10
Unbalance --> ANSWER = Lighter side
Balance --> ANSWER = Weight left outIF all measurements are always equal, then the thirteenth ball is of a different weight.

Clear as mud?

Originally Posted by Rajagra

Are you told which side is heaviest, or just given binary info i.e. same vs not the same?

yes you would know which side is heavier.

Also, can you "cheat" by observing what happens as you remove one ball from each side after each weighing?

I don't think this helps you at all - presume that the difference in weight between the balls is much smaller than their individual weight. But if cheating helps you I'd be interested to know how.

Originally Posted by sidneykidney

Clear as mud?

Yes Sandy, looks like you're right.

My wording (which may be clearer or not) would be:

First split them in three groups A: 4 balls, B:4 balls, C: 5balls
Weighing A vs B you'd narrow it to either the last 5 balls (A=B) or to the first 8 (A!=B).

I.) In case it's in the last 5 balls you weigh 3 of them against any other 3 from the first two groups (which are of the standard weight) and you'll narrow it to either (1) 3 balls and knowing if the odd one is lighter/heavier, or (2) the two balls which you haven't put on the scales yet.
...case (1): weigh any two of these three balls against each other and you're done
...case (2): weigh any one of these two against one of the other balls and you're done again

II.) In case the first measurement showed that the odd ball is among the first two groups (A!=B) take three from group A and two from group B and weigh them against the 5 in group C which now you know are of the standard weight.
...case (1) If they're the same the odd one is among the other three balls that you just left out - one from group A and two from group B. Comparing the two from group B will tell you the answer (you'll also have to use whether in the very first measurement A was lighter or heavier than B).
...case (2) Let's say that in the first measurement group A was lighter than group B (the following logic works the same if it were the other way).
Then if the 5 balls from A&B are lighter than the 5 good ones, the odd one is among the three from group A, and it's lighter than the rest. Otherwise it's among the two from group B and it's heavier. The last measurement will narrow that down exactly as as in case (I).