Donald Cartmill
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Posts posted by Donald Cartmill
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Spoiler
Worst case is 50 saved. The prisoners agree that beginning with #100 an even number ,,every even numbered prisoner will state that his hat is the color right in front of him . I am an odd numbered prisoner ,I know that the prisoner behind me will be calling the color of my hat. It matters NOT how many different colors are in use. The even numbered prisoner will be stating the color of the prisoner directly in front of him, i.e. an odd numbered prisoner will always know his hat color. This is worst case scenario 50 saved. A case could be made ( with only 3 colors in play ) ,whereby statistically 1/3 of the remaining 50 even numbered prisoners will have selected their own hat color as well.
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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.
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21 hours ago, Donald Cartmill said:
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.
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Spoiler
Als' statement is invalid. He therefore is the liar .
Berts' statement is true
Chucks statement is true
Dicks' statement must be true , if there are 3 truth tellers
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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.
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Spoiler
Nails are side by side
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Spoiler
1st person has on a blue hat; 2nd and 3rd persons have on yellow hats
1) 2nd and 3rd persons cannot see a red hat or they could not state 3 poss for their own hats. Therefore each must see a yellow and a blue
2) 1st person has 2 poss ...A) he sees a red and a blue ( void by next statement ); B) he sees two yellow
3) if 1st saw a red ,then 2nd or 3rd would have to see a red ,but since neither 2nd or 3rd can see a red, then poss A) for 1st is void.
4) Therefore option 2) is correct for 1st person i.e. 1st person sees two yellow hats.
5) Since by 1) 2nd and 3rd persons cannot see a red hat and since each have on yellow hats / option B) it then follows that 1st person has on a blue hat
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3 = 4 + 5 - 6 ; 5 - 4 + 5 = 6 ; 3 - 4 + 6 = 5
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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
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Area of white to 3 places is = 0.763
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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
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Spoiler
2R' 1B) BRR ;RBR ; RRB = 3 1R& 2B ) RBB; BRB ;BBR .=3 2R& 1 Y ) YRR ;RYR; RRY; =3 2Y & 1 R ) RYY ; YRY ; YYR ;=3 2B & 1 Y ) YBB ; BYB ' BBY ; = 3 2Y& 1B ) BYY ; YBY ;YYB ;=3 RY&B ) RYB ; RBY ; YRB ; YBR ; BYR ; BRY ;= 6
TOTAL 6 X 3 = 18 PLUS 6 = 24 CORRECTION : in the 1st listed BRR; assuming you are looking at B front and back, then the middle face upward is either R or Y; the same can be said for the R i.e. the upward face must be B or Y; The 2nd R upward is either B or Y .So these possibilities exist reading from left to right R B B; R B Y ; R Y B ; R Y Y YBB ; YBY ; YYB ; YYY = 8 . Each cell or horizontal possibility has a total of 8 poss due to the center face of the cube. Since 1 possible has been counted in the original calculation then 8- 1 = 7. The grand total is 24 x 7 = 168 possibilities
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Spoiler
2 R'S 1B) BRR ;RBR ; RRB = 3 1R& 2B ) RBB; BRB ;BBR .=3 2R& 1 Y ) YRR ;RYR; RRY; =3 2Y & 1 R ) RYY ; YRY ; YYR ;=3 2B & 1 Y ) YBB ; BYB ' BBY ; = 3 2Y& 1B ) BYY ; YBY ;YYB ;=3 RY&B ) RYB ; RBY ; YRB ; YBR ; BYR ; BRY ;= 6
TOTAL 6 X 3 = 18 PLUS 6 = 24
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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
22 minutes ago, Donald Cartmill said: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 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
25 minutes ago, Donald Cartmill said:26 minutes ago, Donald Cartmill said: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
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
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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
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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
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The number is 3 coins
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I am with gavinksong...there is only one answer which is the lowest common denominator for the three groups (plus 1)
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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)
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I get 24 cubes ; 7 in the front face ; 9 in the middle ; and 8 in the rear face
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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%.
Hats of three colors
in New Logic/Math Puzzles
Posted · Edited by bonanova
spoiler added
Proceeding like Izzys' solution. 99 can be saved . # 100 says he can see 35 white hats ; 35 red hats but this does NOT allow him to guess his own color. #99 If he sees the same number of white and red hats knows his hat is black. Proceed in this manner and the only prisoner unable to be sure of color is prisoner #100