Riddles!

[Wildt4lon]

Hang on, they all leave BEFORE the 99th day

[TheGreatErenan]

I suppose Gf!sh's answer would work assuming the gurus all manage to devise the same strategy without speaking to one another.

However, there is another solution:

Suppose there is only one red guru. Then he will see that no other guru has a red mark, and he will deduce that his own mark is red, and therefore he will leave at 4:00 the first day.

Now suppose there are two red gurus. Neither will know if his own mark is red, since they each see one other guru with a red mark. Thus, neither will leave on the first day. However, from the fact that the red guru they can see didn't leave, they will deduce that there must be another red mark. It must be on their own chin. So they would both leave on the second day.

If there are three red gurus, they will not leave on the second day, because they each see two red marks. But from the fact that no one left, they will deduce that there must be a third red mark on their own chin. And they'll leave on the third day.

By induction, if there are n gurus with red marks, they will figure it out and leave on the nth day.

[Wildt4lon]

^That's a blown out version of what I had said

[TheGreatErenan]

(05-12-2013 01:18 PM)Doodat Wrote:  Load 1000, drive 334 miles, drop 666 (not intended),
Go back twice
So 1998 gallons left at 666 miles left
Load 1000, drive 499 miles, drop 501
Load 998, drive 499 miles, drop 499
So 1000 gallons left at 167 miles left
Drive the rest, left with 833 gallons lost 2167

But don't you need some of the gas to go back?

I was going to suggest you drive 200 miles, drop off 600, use the remaining 200 to go back. Repeat until 2000 gallons remain at 200 mile mark. Drive 333 1/3 miles, drop 333 1/3 gallons, go back, load up, drive 333 1/3 miles. Now there's 1000 gallons left at 533 1/3 mile point. Load up all the gas, drive 466 2/3 miles. You save 533 1/3 gallons and waste the rest. Of course, this is assuming you can drive a fraction of a mile using a fraction of a gallon and that gas is necessary to drive back.

[awpertunity]

(05-13-2013 01:48 AM)TheGreatErenan Wrote:  I suppose Gf!sh's answer would work assuming the gurus all manage to devise the same strategy without speaking to one another.

However, there is another solution:

Suppose there is only one red guru. Then he will see that no other guru has a red mark, and he will deduce that his own mark is red, and therefore he will leave at 4:00 the first day.

Now suppose there are two red gurus. Neither will know if his own mark is red, since they each see one other guru with a red mark. Thus, neither will leave on the first day. However, from the fact that the red guru they can see didn't leave, they will deduce that there must be another red mark. It must be on their own chin. So they would both leave on the second day.

If there are three red gurus, they will not leave on the second day, because they each see two red marks. But from the fact that no one left, they will deduce that there must be a third red mark on their own chin. And they'll leave on the third day.

By induction, if there are n gurus with red marks, they will figure it out and leave on the nth day.

Yepppp! I'll stump you eventually....!

[TheGoldenGriffin]

Apparently, his weakness are who am i riddles because no one could guess correctly until i gave hints

[awpertunity]

Haha yeah but since he's a math guy I want to stump him with a math one! How about this...

Two men are paying their tab at a bar and talking about their kids. One man says to the other that he has 3 kids, and the product of their ages is 36 and that the sum of their ages is equal to the cost of their bar tab. The other man says, "Okay.... but I can't figure out how old your kids are from that". The first man agrees and adds, "The oldest one is a boy!" to which the second man replies, "Ah, okay I got it now!"

How old are the three kids?

[TheGoldenGriffin]

Its hard to make up your own riddle, im still thinking

[.Memories.]

(05-13-2013 06:28 AM)awpertunity Wrote:  Haha yeah but since he's a math guy I want to stump him with a math one! How about this...

Two men are paying their tab at a bar and talking about their kids. One man says to the other that he has 3 kids, and the product of their ages is 36 and that the sum of their ages is equal to the cost of their bar tab. The other man says, "Okay.... but I can't figure out how old your kids are from that". The first man agrees and adds, "The oldest one is a boy!" to which the second man replies, "Ah, okay I got it now!"

How old are the three kids?

The first piece of crucial information is the fact that the man says he can't figure it out from what the other man told him. That means that the sum of the three ages has to appear at least twice out of the factors of 36. For example, it can't be 9 * 4 * 1 (= 36) because no other three factors of 36 sum to 14 (= 9 + 4 + 1). If this were the case, he would not need more information from the guy. Therefore, it must be either 9 + 2 + 2 = 13 or 6 + 6 +1 = 13 because they both sum to 13 (and are the only non-unique sums).

The next bit of info, that there is an oldest, finishes it off. There cannot be an "oldest" if there are twins of age 6. Thus, we know the kids' ages are 9, 2, and 2.

[[PETA] Doodat]

36, 1, 1?

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Oh right x.x. But couldn't he have? I mean, if the twin was older by a minute..