a). The kids knew the cookies were Oatmeal?
b). Tilt the drum and pour it until the fluid surface makes a perfect diagonal across the height of the drum
c). The message would take 5 million years to reach the aliens. Although, it's questionable what speed "intergalactic radio message" travels at.
d). A 3 prong plug on the computer vs 2-hole power point from the wall.
I have to say that the questions are not logically watertight enough for an adult... but I'd love it as a kid. I have a friend's 8yr old boy birthday coming up...
I think the proof is simply that the angular velocity of the swimmer is always greater than the cyclops provided he stays within 1/4 of the radius, and he can afford to swim slightly further out with each zig. The swimmer would have to consult trig tables to calculate exactly what angle is optimum at each zig & zag, but in practice the swimmer would just swim more parallel with the shore until the cyclops reversed.
The zigzags would get shallower as the distance from the centre increased. On the whole I think the zigzag method would be more efficient than the swimming in a circle method (and it would have the benefit of p***ing the cyclops off).
And many thanks to you & others for providing these puzzles.
According to my calculations Denise was born 14th May 2002.
Saudi:
The wise man told them that the Sheik meant to race the camels once it arrives? So the slowest camel will be the one most tired?
Billiard:
1. Put 3 on each side - if it balances, then go to step 2.
2. Put the 6 you just scaled aside, divide the other 6 into 2 group of 3 [say, B1 and B2].
2.1 Take B1 and scale against A1 [any 3 balls, all 6 are identical as they balanced]
If B1 and A1 balanced, then the "defective" one must be in B2...
brb...
This is a question that could have many answers obviously, but I think what this question is trying to get at is how the two poor guys could avoid spending the rest of their lives in the desert and still have a chance of winning. Colluding so they pass the line at exactly the same time doesn't count.
This is a question that could have many answers obviously, but I think what this question is trying to get at is how the two poor guys could avoid spending the rest of their lives in the desert and still have a chance of winning. Colluding so they pass the line at exactly the same time doesn't count.
You have 12 billiard balls that look identical in every respect, except one is slightly different in weight (could be heavier or lighter than the other 11). You have a simple balance (one that stays level if both trays contain exactly the same weight, but will tip one way or the other if they are different).
You have to determine by no more than 3 weighings, which ball is different and whether it is heavier or lighter than the others. How can you do it?
Label all 12 balls with [B]Unknown[/B].
Weigh 1.
Put 4 on each side
If they balance then
Label those eight on the balance as C (for Correct weight). We now have 4 Unknowns.
Weigh 2.
Put 3 (of the 4 remaining) Unknowns on one side the balance & 3 of the C on the other.
If they balance then
label the 3 Unknowns on the balance as C.
So we now have 1 Unknowns but we don;t know if it's heavier or lighter, so...
Weight 3a.
Put the only remaining Unknowns of the balance opposite one C, so we can find out if it's heavier or lighter
else if the C and Unknown is heavier then
label the Unknown on the heavier side as [B]UBCBH [/B](Unknown but could be heavier)
label the 2 Unknowns on the lighter side as [B]UBCBL [/B](Unknown but could be lighter)
Weigh 3b
Put one UBCBH and one UBCBL on one side and 2 C on the other
if they balance then
the UBCBL that is not on the scales is the lighter than the other 11
if the side with one UBCBH and one UBCBL is heavier then the UBCBH is heavier than the other 11
else UBCBL is lighter than the other 11
else if the 2 Unknowns are heavier then
label them both UBCBH
Weight 3c.
Label the 2 Unknowns that aren't on the balance as C
Put both the UBCBH on opposite sides of the balance
One will be heavier than the other 11
As they don't balance then
label the 4 not on the balance as Correct
label the heavier 4 Unknowns as UBCBH and the other 4 UBCBL
Weigh 2a.
Put 2 UBCBH and one UBCBL on the LHS of the balance, and 1 UBCBH and 2 UBCBL on the RHS
and leave one of each aside.
if they balance then
Label all 6 on the balance as C
Weight 3d.
Put the remaining UBCBL on the balance opposite a C
if they balance then the remaining one (ie not on the balance) is heavier than the other 11
else the one on the balance (labeled UBCBL) is lighter than the other 11
else they don't balance
Label the 2 that aren't on the balance as C
if the LHS is heavier (it contains 2 UBCBH and only one UBCBL) then
there are 2 possibilities (either the lighter one by itself is lighter than all others, or one of the 2 UBCBH is heavier than all others)
Label the UBCBL on the LHS as C
Label the 2 UBCBH on the RHS as C
We are left with 3 balls - 1 UBCBL and 2 UBCBH
Weight 3e.
Put one of the UBCBH and the only remaining UBCBL opposite 2 C
if they balance then the UBCBH is heavier than the other 1
else if the 2 Cs are heavier the the UBCBL on the balance is lighter than the other 11
else the UBCBH on the balance is heavier than the other 11
else the RHS must be heavier then
just do the inverse of Weight 3e.Not sure what advice the old man gave, but jumping on each others camel & racing to city seems to be the answer. Effectively they are attempting to make the others camel faster.
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