"absolutely useless" is still limited by context unless the request states otherwise.
"I needed something to unclog my drain last night and my iPod was absolutely useless"
A statement you can't disprove even if an ipod has uses elsewhere!
h[/color]i[/color]d[/color]d[/color]e[/color]n[/color] t[/color]e[/color]x[/color]t:[/color] Black can only move either of his two knights, all his pawns are blocked from moving from the 7th row up to the 8th row by black's pieces, only the knights can jump over the pawns. White's knight can...
Hidden text (not a solution but thoughts/questions/spoilers!) are Black King / Queen reversed on purpose? second thought, K/Q switch can't 'legally' happen but board could be 'legal' if black's side is really on the bottom i.e. all pieces somehow switched board sides[/color]
There's a 50-50 chance of winning but the assumption you are risking less than than what you stand to gain is totally wrong. It sounds good, like that riddle with the 3 guys and where's the $2? 50% of the time (when you win) 'what you have to gain is greater than what you stand to lose' is true...
in white:
I believe 1st & 2nd segments of 200 miles (walked 5 times) and 333 1/3 (walked 3 times) are most efficient but don't have a proof
1. start trip with 1000 bananas
2. travel 200 miles, you're left with 800 - stash 600 at 200 mile point, keep 200 for 200 mile trip back.
3. pick up...
(Agreed, but it was the 'nobody one can go alone' line that, for me, ruled out someone crossing either way alone. Hmm, perhaps 'go' may be just mean 'go' and not 'come back'. Two people on the bridge at the same time, even if one's crossing leftward and one's crossing rightward may be legal to...
Why would you assume there is only one flashlight? Post said A flashlight not THE flashlight. If not two would have to cross and then two cross back (if one can't cross alone one shouldn't be able to cross back alone) and all 4 would never all get across
rachmaninoff,
Case 1 is clever I didn't think of poisoning everyone to come up with the antidote! I'm not following case 2 yet though, if no one died all I can see is the poison was not (0,0,0,...0) but it still could be any other bottle. Any 0 digits in the number of the poison can just also...
number the bottles 0-1023 in binary numbers (10 digits), each bottle number with first digit a "1" is drunk by person 1, second digit a "1"? then person 2 drinks some, etc. etc. If say, the 1st 4th and 9th die then then bottle was 1001000010 (decimal 578)
It's Aditya, he's the one who made up the riddle!
I agree with Gale17 on the dog, once he's in the woods he's running IN the woods and not INTO the woods.