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Originally posted by sonnny1
reply to post by spy66
Am I A Bipolar Opposite?
Originally posted by sirric
Can you move one matchstick and still have an equation that is correct? I think there are at least three answers – can you find all three?
LINK
3-9=6
3+3=6
not sure of the last one
That's a good one.
Here is a math logic question
3 men decide to rent a room for the night, The room is $30
They each pay $10 and go up to the room.
The manager made a mistake and realizes that the room only cost $25 for the night, so he gives the bellboy $5 to give back to the 3 men.
As the bellboy was in the elevator, he knew that he couldn't split $5 between 3 people so he pocket $2 and gave each man $1 back.
How much did each man pay for the room?
Where did the missing dollar go?
sirricedit on 30/12/12 by sirric because: (no reason given)
Solution:
3,3,8
Lets break it down. The product of their ages is 72. So what are the possible choices?
2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19
The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.
2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14
Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.
Originally posted by john_bmth
Originally posted by Logarock
reply to post by john_bmth
If you get a yes answer first it has to be from the random God.
edit on 31-12-2012 by Logarock because: n
50% of the time the random god will answer "no".
Originally posted by Ryanssuperman
Originally posted by smithjustinb
reply to post by sirric
Each man payed 9 dollars for the room. There is no missing dollar.
9x3 = 27 + 2 (In the bellboys pocket) = 29. Therefor, there's a "missing" dollar.edit on 31-12-2012 by Ryanssuperman because: (no reason given)
"Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language in which the words for 'yes' and 'no' are 'da' and 'ja', in some order. You do not know which word means which."
Originally posted by intrptr
reply to post by BigBrotherDarkness
Lol. My solution was simpler. Shoot one in the head and wait for the responses from the others...
You don't have to say anything. The remaining two will sing like birds.
Edit:I have an old one: You live in a house where all four walls face south. A bear walks by, what color is the bear?edit on 31-12-2012 by intrptr because: (no reason given)
Originally posted by john_bmth
reply to post by Logarock
Using your approach there are many sequences (where the random god isn't first) with multiple god orderings valid for each sequence. Ergo, it is not a solution to the challenge.edit on 31-12-2012 by john_bmth because: (no reason given)
Originally posted by Logarock
Originally posted by john_bmth
reply to post by Logarock
Using your approach there are many sequences (where the random god isn't first) with multiple god orderings valid for each sequence. Ergo, it is not a solution to the challenge.edit on 31-12-2012 by john_bmth because: (no reason given)
No kidding? Many? There are only 3 sets 2 sequences, 6 total sequences. One set is already figured out for you. Did you fail alegbra?edit on 31-12-2012 by Logarock because: n
Originally posted by john_bmth
Originally posted by Logarock
Originally posted by john_bmth
reply to post by Logarock
Using your approach there are many sequences (where the random god isn't first) with multiple god orderings valid for each sequence. Ergo, it is not a solution to the challenge.edit on 31-12-2012 by john_bmth because: (no reason given)
No kidding? Many? There are only 3 sets 2 sequences, 6 total sequences. One set is already figured out for you. Did you fail alegbra?edit on 31-12-2012 by Logarock because: n
There are 24 answer sequences using your approach. Your solution is valid for four, the rest are indeterminate. It is not a valid solution because there are 20 sequences it doesn't solve. But by all means explain how your solution is valid when it doesn't actually solve the challenge.edit on 1-1-2013 by john_bmth because: (no reason given)