Let's Have a Funny Pic Thread! Mk 35

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mongooz said:
lotta Huns in baltimore, are there?
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Calling people "Hon" is a big thing in certain parts of Baltimore. John Waters made a big thing of the "Hon" culture; allegedly paired with beehive hairdos, ala "Hairspray".
There is a restaurant there called Cafe Hon; the owner pissed people off by attempting to trademark "Hon". (By the way, the food there was nothing to brag about as of sometime around 2005.)
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mongooz said:
this whole godamn year has been one "hold my beer" cluster fuck after another.
everyone needs to stop saying "could this get any worse".....seriously!!! :mad:
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There's a classic Israeli sketch about a boy who grew up in a shabby village.
"Cheer up, son", his parents said, "things could be a whole lot worse"
So he cheered up, and things really got a whole lot worse.
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shoeless said:
Might want to double check that list. You may have missed some possible combinations...and when I say "may have"...well, you did. :wink: You're on the right track though, and once you get the complete list, the answer will reveal itself. And yes, integer yearly age only.
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OK, here's the complete list. B1 and B2 are the younger brothers. B3 is the oldest. Again, I'm assuming that B3's age expressed as an integer must be greater than B1 and B2's age as an integer.

B1 = 1, B2 = 1, B3 = 36, S = 38
B1 = 1, B2 = 2, B3 = 18, S = 21
B1 = 1, B2 = 3, B3 = 12, S = 16
B1 = 1, B2 = 4, B3 = 9, S = 14
B1 = 2, B2 = 1, B3 = 18, S = 21
B1 = 2, B2 = 2, B3 = 9, S = 13
B1 = 2, B2 = 3, B3 = 6, S = 11
B1 = 3, B2 = 1, B3 = 12, S = 16
B1 = 3, B2 = 2, B3 = 6, S = 11
B1 = 3, B2 = 3, B3 = 4, S = 10
B1 = 4, B2 = 1, B3 = 9, S = 14

Maybe I'm dense, but the answer hasn't revealed itself.
 
GomezAddams said:
OK, here's the complete list. B1 and B2 are the younger brothers. B3 is the oldest. Again, I'm assuming that B3's age expressed as an integer must be greater than B1 and B2's age as an integer.

B1 = 1, B2 = 1, B3 = 36, S = 38
B1 = 1, B2 = 2, B3 = 18, S = 21
B1 = 1, B2 = 3, B3 = 12, S = 16
B1 = 1, B2 = 4, B3 = 9, S = 14
B1 = 2, B2 = 1, B3 = 18, S = 21
B1 = 2, B2 = 2, B3 = 9, S = 13
B1 = 2, B2 = 3, B3 = 6, S = 11
B1 = 3, B2 = 1, B3 = 12, S = 16
B1 = 3, B2 = 2, B3 = 6, S = 11
B1 = 3, B2 = 3, B3 = 4, S = 10
B1 = 4, B2 = 1, B3 = 9, S = 14

Maybe I'm dense, but the answer hasn't revealed itself.
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Keep in mind that order doesn't matter. That is, 2 x 3 x 6 is the same as 3 x 2 x 6, so there is no reason to have repeats like that in your set. You are still missing one in your set, however. So if we only look at unique combinations we have,

1x1x36, S=38
1x2x18, S=21
1x3x12, S=16
1x4x9, S=14
1x6x6, S=13
2x2x9, S=13
2x3x6, S=11
3x3x4, S=10

Does that help?
 
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