Kogorous Riddle Thread and eworms Deduction Thread

[c-square]

1) if a box contain exactly one diamond he can take it

Do you mean he can take the diamond, leaving the box empty to be reused, or do you mean he can take the box, leaving one less box available?

[Ankor]

1) if a box contain exactly one diamond he can take it

Do you mean he can take the diamond, leaving the box empty to be reused, or do you mean he can take the box, leaving one less box available?

Well it originaly it meant that he could take diamond without putting it to an other box, leaving box empty. But as number of boxes is greater than number of diamonds this is not a principal question.

[VQ]

2) put all the diamonds from one box to any other box

So, can we take all the diamonds, let's say from the box with 69 and put them in the box with 35?
[spoiler]If the answer is yes, then I would say the thief can't steal anything.[/spoiler]

[Ankor]

So, can we take all the diamonds, let's say from the box with 69 and put them in the box with 35?
[spoiler]If the answer is yes, then I would say the thief can't steal anything.[/spoiler]

Yes we can.

But why do You think answer is none?
As without explanation it would look like an guessing 

[VQ]

So, can we take all the diamonds, let's say from the box with 69 and put them in the box with 35?
[spoiler]If the answer is yes, then I would say the thief can't steal anything.[/spoiler]

Yes we can.

But why do You think answer is none?

[spoiler]The number of diamonds inside the 3 boxes are all odd numbers 35, 69 and 91. No matter how I add or divide them I can't seem to get the number needed to divide the diamonds to get 1 diamond inside various boxes, which I believe is the number 128 or one of his divisors:
128/2=64
64/2= 32
32/2= 16
16/2= 8
8/2=4
4/2= 2
2/2=1

I did find a solution by removing one diamond though.
Are you sure those are the right numbers of diamonds? or Am I not seeing something?[/spoiler]

[Ankor]

So, can we take all the diamonds, let's say from the box with 69 and put them in the box with 35?
[spoiler]If the answer is yes, then I would say the thief can't steal anything.[/spoiler]

Yes we can.

But why do You think answer is none?

[spoiler]The number of diamonds inside the 3 boxes are all odd numbers 35, 69 and 91. No matter how I add or divide them I can't seem to get the number needed to divide the diamonds to get 1 diamond inside various boxes, which I believe is the number 128 or one of his divisors:
128/2=64
64/2= 32
32/2= 16
16/2= 8
8/2=4
4/2= 2
2/2=1

I did find a solution by removing one diamond though.
Are you sure those are the right numbers of diamonds? or Am I not seeing something?[/spoiler]

Numbers are correct.
And well if one would see the 'similarity' between non-empty boxes after first step, then one could give an answer straight away.

[c-square]

Numbers are correct.
And well if one would see the 'similarity' between non-empty boxes after first step, then one could give an answer straight away.

Okay...

[spoiler]As VQ pointed out, to be able to get to 1 diamond, we need to get a box with an amount of diamonds that is a power of 2.  Or to put it another way, only has 2 as a prime factor.

Let's take a look at the possible first steps:

Scenario 1: Add the 69 and 91 boxes together

That leaves us with two boxes with 160 and 35 diamonds respectively.

The prime factors of 160 are 2 and 5.  If we could get rid of that 5, we'd be able to get to our power of 2.  The problem is that 35 has prime factors of 5 and 7.  And whenever you add a number with a prime factor of 5 to another number with a prime factor of 5, the result is a number with a prime factor of 5.  Similarly, any time you divide an even number with a prime factor of 5 by 2, the result has a prime factor of 5.  So there's no way to get rid of that prime factor of 5, so in this scenario, the thief would end up with 0 diamonds.


Scenario 2: Add the 35 and 69 boxes together

That leaves us with two boxes with 104 and 91 diamonds respectively.

The prime factors of 104 are 2 and 13.  The prime factors of 91 are 7 and 13.  The same problem exists as above.  No matter how he adds the boxes together or divides by 2, he'll never be able to get rid of that 13 prime factor, meaning he leaves empty handed.


Scenario 3: Add the 35 and 91 boxes together

That leaves us with two boxes with 126 and 69 diamonds respectively.

The prime factors of 126 are 2, 3 and 7.  The prime factors of 69 are 3 and 23.  Because they share the prime factor 3, any combination of adding together or dividing by 2 will also have that prime factor, making it impossible to get to an amount of diamonds that is a power of 2.  Again, the thief gets nothing.

So, according to the above, the maximum number of diamonds the thief can take is zero.  Am I missing something??[/spoiler]

[kidSherlock]

[spoiler]As VQ pointed out, to be able to get to 1 diamond, we need to get a box with an amount of diamonds that is a power of 2.  Or to put it another way, only has 2 as a prime factor.

Let's take a look at the possible first steps:

Scenario 1: Add the 69 and 91 boxes together

That leaves us with two boxes with 160 and 35 diamonds respectively.

The prime factors of 160 are 2 and 5.  If we could get rid of that 5, we'd be able to get to our power of 2.  The problem is that 35 has prime factors of 5 and 7.  And whenever you add a number with a prime factor of 5 to another number with a prime factor of 5, the result is a number with a prime factor of 5.  Similarly, any time you divide an even number with a prime factor of 5 by 2, the result has a prime factor of 5.  So there's no way to get rid of that prime factor of 5, so in this scenario, the thief would end up with 0 diamonds.


Scenario 2: Add the 35 and 69 boxes together

That leaves us with two boxes with 104 and 91 diamonds respectively.

The prime factors of 104 are 2 and 13.  The prime factors of 91 are 7 and 13.  The same problem exists as above.  No matter how he adds the boxes together or divides by 2, he'll never be able to get rid of that 13 prime factor, meaning he leaves empty handed.


Scenario 3: Add the 35 and 91 boxes together

That leaves us with two boxes with 126 and 69 diamonds respectively.

The prime factors of 126 are 2, 3 and 7.  The prime factors of 69 are 3 and 23.  Because they share the prime factor 3, any combination of adding together or dividing by 2 will also have that prime factor, making it impossible to get to an amount of diamonds that is a power of 2.  Again, the thief gets nothing.

So, according to the above, the maximum number of diamonds the thief can take is zero.  Am I missing something??[/spoiler]

[spoiler]yes you are missing something,
you forgot to add all of the diamonds in the 3 boxes.

35 + 69 + 91 = 195
prime factors are

5x3x13

so there are still no diamonds that the thief could steal,

P.S. That thief is really dumb,,[/spoiler]

[Ankor]

Yep, thief can steal 0 diamonds.
To bad for him.

And perfectly correct explanation by c-square.

And I am sorry maybe riddle was not very good - as subconsciously it is assumed that there should be a way to get at least 1 diamond and 0 is wrong answer. (which is why personally I spent more than couple of days trying to find correct answer first time I seen it  )

yes you are missing something,
you forgot to add all of the diamonds in the 3 boxes.

35 + 69 + 91 = 195
prime factors are

5x3x13

so there are still no diamonds that the thief could steal,

P.S. That thief is really dumb,,

Well this thing is redundant as after any step taken first it is seen that there is now way thief will steal anything, so second step (putting diamonds into one box) is not necessary.

[c-square]

Yes, that was a strange one, because you're first instinct is to assume it's doable.

Palindrome time! !emit emordnilaP

What the chairwoman shouted to the city councilman who had fallen asleep during a council session:

"_ _ _ _  _ _  _ _ _ _ ,  _ _ _ !"

[kidSherlock]

i think i remember this one
but because i don't have any riddle i'll pass,

[eworm]

But as alarm only is only partly disarmed thief can do only the following things with the safe without being caught:
1) if a box contain exactly one diamond he can take it
2) put all the diamonds from one box to any other box
3) put exactly half of the diamonds in the box to any other box (if diamond amount in the box is not even this cannot be done).

I love this security system. The thief has an access to all the boxes and yet ends up with nothing ^^

What the chairwoman shouted to the city councilman who had fallen asleep during a council session:

"_ _ _ _   _ _   _ _ _ _ ,   _ _ _ !"

I think I may have the first and the last word... But what's in between, I can't figure out. Maybe I'm wrong after all...
[spoiler]"Rise to vote, sir!" ?[/spoiler]

Anybody thinking about Kogorou's case?

[c-square]

What the chairwoman shouted to the city councilman who had fallen asleep during a council session:

"_ _ _ _  _ _  _ _ _ _ ,  _ _ _ !"

I think I may have the first and the last word... But what's in between, I can't figure out. Maybe I'm wrong after all...
[spoiler]"Rise to vote, sir!" ?[/spoiler]

Correct!!  Good job, eworm!  Your turn. 

[VQ]

Yep, thief can steal 0 diamonds.
To bad for him.

[spoiler]I'm pretty sure I said first the thief couldn't steal anything and gave an explanation, but whatever.

About Kogorou's case, someone already solved Aunt Merry's cards?[/spoiler]

[Silver1412]

Come on eworm! Hurry up will you ~nya? I'm back and I want a good riddle! Stop making those comics! Nevermind the fact they're hilarious...