Donald Cartmill

9 minutes; i.e. 1 rower & 2 non rowers cross in 4 minutes ; 1 rower returns in 1 minute ; 2 rowers and 1 non rower cross in 4 minutes.

Not exactly a proof ,but if all soldiers watch the soldier to their left, then the soldier to the extreme right is NOT being watched. and the reverse is true. Taking the smallest odd number 3, if the middle man looks left then the man on the right is not watched . The only way all ARE watched is if they are in a circle.

3 = 4 + 5 - 6 ; 5 - 4 + 5 = 6 ; 3 - 4 + 6 = 5

Area of the white is = 1 (1/4) + 9(1/4 x 1/4 ) + 27 ( 1/4 x 1/4 x 1/4) + 81 ( 1/4 x 1/4 x 1/4 x 1/4 ) The lowest common multiple of all of the divisors is 1024 or 4 x 4 x 4 x 4 Area of the white is = 256/1024 + 192 / 1024 +144/1024 +108/1024 + 81/1024; Totaling these =781 / 1024 then dividing 1024 into 781 you get 0,762695313 rounding off to 3 places = 0.763

Area of white to 3 places is = 0.763

remove 1 match from the 6 making it a 5 ; Then place the match horizontally to the top of the 1 ,making it a 7; 5 + 7 + 13

Tried to edit my previous answer and lost everything . Correction ...Red sq "beneath "gray and rotated 60 degrees; Red sq "a top" gray sq and rotated 30 degrees new answer is 8 ; 1 beneath the gray but rotated 60 degrees ,a 2nd above the gray but rotated 30 degrees.4 more one at each corner= 6 ,however if you rotate the top two corner reds away from center you can place a red with one corner touching the gray at the mid point of the top side of the gray. You can do the same at the bottom two corner "Reds",rotating them outward and then placing the 8th red with a corner touching the gray at the midpoint of the grays' bottom side. If the red over gray is considered to be touching the red beneath the gray then the answer is 7

Captain Ed had the answer a red beneath the gray but rotated 60 degrees; another red a top of gray and rotated 30 degrees. Then you have one red at each of the four corners,which totals 6 new answer is 8 ; 1 beneath the gray but rotated 60 degrees ,a 2nd beneath the gray but rotated 30 degrees.4 more one at each corner= 6 ,however if you rotate the top two corner reds away from center you can place a red with one corner touching the gray at the mid point of the top side of the gray. You can do the same at the bottom two corner "Reds",rotating them outward and then placing the 8th red with a corner touching the gray at the midpoint of the grays' bottom side. If the red over gray is considered to be touching the red beneath the gray then the answer is 7

Actually I think this goes back a long way to the old Saturday evening post. The military had made a study of all of the planes that had made it back from air raids. Cataloged every bullet hole that had pierced the hull of these airplanes . There were suggestions that these areas should be armor plated . Someone else spoke up making the statement, that in fact these were the planes that made it back meaning these areas were less damaging to the plane ability to fly ,and the armor plating could best be used else where

The number is 3 coins

I am with gavinksong...there is only one answer which is the lowest common denominator for the three groups (plus 1)

The maximum number of coins for the solution is 3. This is basically what you arrive at when deriving the odd coin in a group of 12 ,allowing only 3 weighings, the maximum number for a solution is 3. Bottom line is that step 1) taking 1/3 of the coins on each pan and leaving 1/3 off. If they balance then step 2) divide those previously left off again into 1/3 on each pan and 1/3 off. 3) You will continue this until you are down to 3 coins. Then 1 on each pan , if they balance then the one left off is the odd coin. A) Now if at any time the number of coins is not evenly divisible by 3 ,then the left over coins go in the pile left off. B) If at any time during the early stages the pans do not balance , you will remove the right pan coins and divide the left pan into half placing 1/2 on each pan...they balance ,then the odd is obviously in the group removed from the right pan and start again at step 2) Now if they do not balance repeat B)

I get 24 cubes ; 7 in the front face ; 9 in the middle ; and 8 in the rear face

50% chance of more heads by (A) ; " X " equal number of coins by (A) and (B) are 50/50 , Therefore the results of a single coin toss is the answer ,which is 50%.