Donald Cartmill

While the combined speed ratios provide the correct answer . I still maintain the speed ratio of A to B is not crucial to the solution. i.e. Let us suppose that A is merely to step out at the finish line to congratulate B and C the competitors in a two man race . A steps out when B has run 80 meters; Congratulates B when B completes the race and notes at the time C has completed 80 meters. Where was C when A first made his appearance ? The answer is 4/5 x 80 = 64 meters While the combined speed ratios provide the correct answer . I still maintain the speed ratio of A to B is not crucial to the solution. i.e. Let us suppose that A is merely to step out at the finish line to congratulate B and C the competitors in a two man race . A steps out when B has run 80 meters; Congratulates B when B completes the race and notes at the time C has completed 80 meters. Where was C when A first made his appearance ? The answer is 4/5 x 80 = 64 meters To beat a dead horse let A step out when B is at 85 meters; He then congratulates B as he finishes the race and notes C is at 80 meters. He wonders where was C , when I first stepped out ??? The answer is of course 4/5 x 85 = 68 meters

Yes ,I agree Capt Eds' answer is not quite as straight forward as could be. After A reaches 100 m ...20 m ahead of B , the speed of A is not significant to the solution. The key is that when B is at 100 ,C is at 80. Therefore C is 8/10 the speed of B and it follows that when B was at 80 m...C was at 8/10 x 80 = 64 m. A is at 100 m when C is at 64 therefore A beats C by 36 m

A crosses finish 20 meters a head of B ; Now B crosses 20 meters ahead of C ; i.e. B at 100 meters ...C at 80 meters; therefore C was traveling at 8/10 the speed of B. Therefore when B was at 80 meters ...C was 8/10 x 80 meters or 64 meters. Thus A beat C by 36 meters

The answer given is 5! , which closely resembles 51 or 5 ! exclamation mark. It is however 5 factorial or 5 x 4 x 3 x 2 x 1= 120 25 - 55 + ( 85 + 65 ) = 120 =5 ! 25 - 55 + (150 ) ; -55 + 175 ; 175 - 55 == 120

the odds of eating a poison apple for the 1st bowl = 3/5 + 3/4 = 12/20 + 15/20 = 27/20 = !.35 chances of eating the P/A 2nd bowl =2/5 + 2/4= 2/3 = 24 /60 +30/60 + 40/60 = 84/60 = 1.4 CHANCES Therefore you would eat from the 1st bowl with a 0.05 better chance of avoiding the P?A

I'm not sure I recognize your solution 6 from its description of how it's constructed, but your final description of it seems right. Nice solve. Let me attempt to clarify 6) 4 points A,B,C,D, lay on two intersecting circles having the same radius. point A is the center of one circle with C and D laying on its circumference , D is the center of the 2nd circle with A and B laying on its circumference. The 4 points create a trapezoid in which The diagonals AC , & BD plus the base AD are equal all being a radius. Points B and C lay on their respective circumferences at points where the chords AB = CD = equals the top BC I don't know if that is any more clear

4 would be the smallest number. assign a unit weight of 1 ;2 ;3 ;4; to the 4 weights place units 1 and 4 on left pan and 2 and 3 on the right pan each weigh 5 and balance sorry I misunderstood the question in my earlier answer

Two of you got the answer ,but seemingly struggled with the simplicity of the solution. i.e. If a second man started up the mountain at the same time the other was coming down...regardless of their relative speeds ,stopping to rest , at some point they must meet , and obviously at the same time.

A holy man starts up a mountain at 7:00 am ; Arrives at the top at 7:00 pm He goes thru some worship ceremony and the next morning ,and the next morning at 7:00 am he starts back down the mountain. He arrives at the bottom at 7:00 pm . Is there a place where going up and coming down he was at the place at the same time both days. Proof please

First thought is that no matter how fast Albert runs his front always sees 2000 rain drops. i.e. his front is 2 square meters and he is seeing 1000 drops/cu meter . On the other hand the top of Albert 0.20 square meters. if Albert was 1 sq meter, moving at 1 m/sec ,then his top sees ( 1,000 / cu meter falling at 10 m/sec = 10,000 drops) however ,Albert has only a 0.20 sq meter profile therefore his top would see ( 10,000 x 0.20 = 2,000 drops ) if he moves at 2 meters / sec then he is hit on top by 1,000 drops. if then he is moving at 4 m/sec 2,000 divided by 4 = 500 drops. There fore Albert should run as fast as possible

First thought is that no matter how fast Albert runs his front always sees 2000 rain drops. i.e. his front is 2 square meters and he is seeing 1000 drops/cu meter . On the other hand the top of Albert 0.20 square meters. if Albert was 1 sq meter moving at 1 m/sec ,then his top sees ( 1,000 / cu meter falling at 10 m/sec = 10,000 drops ) however his profile is only 0,20 sq meters and therefore he sees ( 0.20 x 10,000 drops = 2,000 drops ); If he moves at 2 m/sec ; then he sees 1,000 drops at 4 m/sec he would see 500 drops,etc So Albert should run as fast as possible.

understand that even in the hashtag ,the matches were / are not in the actual same plane , then the following will result in 12 right angles. Two of the matches are crossed giving us 4 right angles. The 3rd match is vertical to the plane of the other two ,and extends equidistant above and below the plane of the other two matches. The vertical portion above the plane makes 4 right angles with the two crossed matches. The vertical portion below the plane makes 4 more right angles Total 12

This is for you guys who are having trouble with the real hard stuff. A farmer has a barrel of pickles weighing 100 pounds . Pickles are 99% water . He leaves the lid off of the barrel until the pickles are 98% water. How much does the barrel of pickles now weigh. OK experts give the little guy a chance

I'd play Tic-Tac-Toe on a magic square: 2 7 6 9 5 1 4 3 8 If I get to go first, I'll start with the "middle square" and choose 5. Next I'll choose a corner -- 2, 4, 6, or 8 -- depending on what you pick. If you chose one of the edges, I choose a corner adjacent to it; if you chose a corner, I choose a neighboring corner. Presumably, your next choice will be forced so as to prevent me from scoring 15 on a diagonal, after which I can guarantee a win for myself by choosing another corner number, thus leaving me with two ways to win on my next turn. Your solution is invalid. Your opponent chooses a 4; you choose a 2 or an 8; opponent chooses either 2 or 8; You are now forced ,you cannot take the remaining corner or your opponent score a 15 across the bottom ;if your 2nd was a 2; Therefore you must go a 3 to block ; he then must go 7 to block ....No winner

You have 12 balls of equal size and weight ,except one ball is either lighter or heavier than the others. Using an analytical balance only 3 times ,identify the odd ball and weather it is hvy or light

Randomly placed is NOT sufficient info to establish the % of unbounded belt on the in going (top side ) half of the belt . Also while implied there is no mention of any belt length segment Therefore the avg will b.e over 50%. How much over 50%. Assuming each piece of luggage as placed on belt was 2 ft wide. 2nd piece is 1 inch from 1st, and the 3rd is 2 inches away; Then the 4th must be at least 3 inches away from the 3rd and becomes the 1st in the next segment of luggage . if this were to repeat on and on, then the % of unbound belt would be 3 inches divided by the total length of the segment. i,e, (24 +1 + 24 + 2 + 24 +3 ) = 78; Therefore 3 /78 =0.03845 or 3.8% for the top belt and in the scenario added to the 50% = 53.8 % This would be the lowest % using a full inch as a minimum unit.

Good work straight to the point ...I screwed around and came up with the same Q ,but a lot longer getting there Kleene: Is the total value of the coins 15? There are currently 3 possibilities for their total: 9, 15, and 21. If the answer to the question is yes, everybody knows the total. If the answer to the question is no, one two of the three (both Carroll and Kurt) should be able to deduce whether the total is 9 or 21. If anyone has 7, 8, or 9 coins, they can deduce that there must be 21 coins (they would have to total more than 9). If anyone has 1, 2, or 3 coins, they can deduce that there must be 9 coins (they couldn't sum to 21 given the rules). Since Kleene has 5 coins, there is no way to make 21 coins without someone having 7 and the other having 9. There's no way to make 9 coins without someone having 1 and the other having 3. Small correction ; for 21 no 8 possible ; The same for 9 i.e. no 2 possible

I have it ...the 6th solution 1) a square with 4 sides and two diagonals =6 lines 2) would be a diamond shape created by two equilateral triangles; 5 sides equal and one long diagonal 3) An equilateral triangle ,3 equal sides ; The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter 4) an equilateral triangle is 3 lines connecting points "A" ,"B" and "C", Assume "A" is the vertex and "B" and "C" are the base points. A 4th point "X"gives us a line the same length as a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C"; This arrangement gives us 4 lines of equal length . Now the lines connecting the 2 base points "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines 5) an equilateral triangle is 3 lines connecting points "A" ,"B" and "C", Assume "A" is the vertex and "B" and "C" are the base points. A 4th point "X"gives us a line the same length as a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC = 4 lines of equal length; 2 short lines BX and CX 6) 4 points A,B,C,D, lay on two intersecting circles. point A is the center of one circle C and D lay on this circle. D is the center of the 2nd circle with A and B laying on the circle. The 4 points create a trapezoid in which the distance BC , is the same length as chords AB and CD (that is 3 short lengths ) The diagonals AC ,AD and BD are equal all being equal to the radius. That should be all of them ???? 6) 4 points A,B,C,D, lay on two intersecting circles having the same radius. point A is the center of one circle C and D lay on this circle. D is the center of the 2nd circle with A and B laying on the circle. The 4 points create a trapezoid in which the distance BC , is the same length as chords AB and CD (that is 3 short lengths ) The diagonals AC ,AD and BD are equal all being equal to the radius.

6 solutions with number 6 being only theoretically correct

If I understand your descriptions, case 7 is also case 4; case 6 is also case 5. Can you check that?

Spoilers please