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night_shift
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Re: mathematical brain teaser thread *DELETED* [Re: extreme]
#22002408 - 07/26/15 11:34 PM (8 years, 6 months ago) |
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Post deleted by night_shift
Reason for deletion: [delete]
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extreme


Registered: 04/05/11
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Re: mathematical brain teaser thread [Re: night_shift]
#22002525 - 07/27/15 12:28 AM (8 years, 6 months ago) |
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No M2 is a 2nd magician who is standing off to the side while M1 and the audience member choose the cards.
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secondorder
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Re: mathematical brain teaser thread [Re: extreme]
#22002781 - 07/27/15 03:13 AM (8 years, 6 months ago) |
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Quote:
1. Weigh A against B. Let's say A > B. 2. Weigh C against D. Let's say C > D. 3. Weigh A against C. Let's say A > C. We now know that A > C > D and B and E are somewhere there in between. (E could be the heaviest still.) 4. Weigh E against C. If E > C, 5. weigh E against A. If E > A, then E > A > C > D (case 1). If E < A, then A > E > C > D (case 2). If E < C, 5. weigh E against D. If E > D, then A > C > E > D (case 3). If E < D, then A > C > D > E (case 4). We now need to place B somewhere and we know that A > B. Depending on which of the four cases we are in, we weigh B against the object that is lighter than the heaviest lighter than A (case 1: D, case 2: C, case 3: E, case 4: D). This is the sixth weigh. If B is heavier we weigh it against the object on the left (case 1: C, case 2: E, case 3: C, case 4: C). If B is lighter we weigh it against the object on the right (case 1: no need, we're done, case 2: D, case 3: D, case 4: E). This is the seventh weigh and in all cases we can determine the order of all five objects.
Nicely Done!
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DividedQuantum
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Re: mathematical brain teaser thread [Re: night_shift]
#22003583 - 07/27/15 09:39 AM (8 years, 6 months ago) |
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Quote:
night_shift said:
Quote:
DividedQuantum said: PROBLEM #21
Quote:
Five objects, no two the same weight, are to be ranked in order of increasing weight. You have available a balance scale, but no weights. How can you rank the objects correctly in no more than seven separate weighings?
1. Weigh A against B. Let's say A > B. 2. Weigh C against D. Let's say C > D. 3. Weigh A against C. Let's say A > C. We now know that A > C > D and B and E are somewhere there in between. (E could be the heaviest still.) 4. Weigh E against C. If E > C, 5. weigh E against A. If E > A, then E > A > C > D (case 1). If E < A, then A > E > C > D (case 2). If E < C, 5. weigh E against D. If E > D, then A > C > E > D (case 3). If E < D, then A > C > D > E (case 4). We now need to place B somewhere and we know that A > B. Depending on which of the four cases we are in, we weigh B against the object that is lighter than the heaviest lighter than A (case 1: D, case 2: C, case 3: E, case 4: D). This is the sixth weigh. If B is heavier we weigh it against the object on the left (case 1: C, case 2: E, case 3: C, case 4: C). If B is lighter we weigh it against the object on the right (case 1: no need, we're done, case 2: D, case 3: D, case 4: E). This is the seventh weigh and in all cases we can determine the order of all five objects.
  
Correct! Excellent!
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DividedQuantum
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PROBLEM #22
Quote:
How many different ten digit numbers, such as 7,829,034,651, can be written using all ten digits?
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ballsalsa
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this one can be done with a button on the calculator...
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deff
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3,265,920 ?
assuming the number can't start with zero then it's
9 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 3265920
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DividedQuantum
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Re: mathematical brain teaser thread [Re: deff]
#22003658 - 07/27/15 09:56 AM (8 years, 6 months ago) |
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Quote:
deff said: 3,265,920 ?
assuming the number can't start with zero then it's
9 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 3265920
  
Correct!
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DividedQuantum
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PROBLEM #23:
Quote:
Two ferryboats start at the same instant from opposite sides of a river, traveling across the water on routes at right angles to the shores. Each travels at a constant speed, but one is faster than the other. They pass at a point 720 yards from one shore. Both boats remain in their slips for 10 minutes before starting back. On the return trips they meet 400 yards from the other shore.
How wide is the river?
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Rhizoid
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Quote:
DividedQuantum said: PROBLEM #23:
The river is 1280 yards wide.
Let w be the width of the river. Let v be the speed of the first ferry. Let u be the speed of the second ferry.
The slip time can be ignored since it is the same for both boats.
They first meet at time t0: v*t0 = w - u*t0 = 720 This gives us t0 = 720/v and t0 = w/(v+u)
Second meeting is at time t: 2w - vt = ut - w = 400 3w = (v+u)*t t = 3w/(v+u) = 3*t0 = 3*720/v = 2160/v
Plug t into second meeting: 400 = 2w - v*(2160/v) = 2w - 2160 400 + 2160 = 2w w = 1280
Edit: sorry this is wrong, I forgot that the 400 position is from the other shore, so it should be w-400 instead. So the final equation is w-400 = 2w - 2160 which gives w = 2160-400 = 1760 yards.
Edited by Rhizoid (07/29/15 06:07 PM)
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DividedQuantum
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Re: mathematical brain teaser thread [Re: Rhizoid]
#22016359 - 07/29/15 06:13 PM (8 years, 6 months ago) |
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Correct! One mile wide. Well done!
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DividedQuantum
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PROBLEM #24
Quote:
Oil is found at an underground spot 21,000 feet from one corner of a rectangular farm, 18,000 feet from the opposite corner, and 6,000 feet from a third corner. How far is it from the fourth corner?
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ballsalsa
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i drew a rectangle. the oil is at the corner of a right triangle with one leg measuring 6000ft, and the other measuring 21000ft. i solved for the hypotenuse which corresponds to one long side of the rectangle. the long side measures 21840.33 ft. I subtract 18000 from the 21840.33 which leaves 3840.33.
therefore, the distance from the fourth corner is 3840.33ft
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DividedQuantum
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Re: mathematical brain teaser thread [Re: ballsalsa]
#22016771 - 07/29/15 07:33 PM (8 years, 6 months ago) |
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I'm sorry, that is incorrect.
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secondorder
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I drew a rectangle played around with the point and length until I got the same ratios as mentioned above. 2cm from one corner (6000ft) 6cm from another (18000ft) and 7cm from the opposite of the 6cm corner (21000ft). I then measured from this point to the remaining corner; it was 9cm. Keeping the ratio of 1cm to 3000ft, that means it is 27000ft from the remaining corner.
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DividedQuantum
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Re: mathematical brain teaser thread [Re: secondorder]
#22017171 - 07/29/15 08:47 PM (8 years, 5 months ago) |
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DividedQuantum
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PROBLEM #25
Quote:
During a baseball game in Mudville, Casey was Mudville's lead-off batter. There were no substitutions or changes in the batting order of the nine Mudville men throughout the nine-inning game. It turned out that Casey came to bat in every inning. What is the least number of runs Mudville could have scored?
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DisoRDeR
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Casey always gets out and ends innings, except the first.
1st inning -- 3 at bats, 0 runs 2nd inning -- 7 at bats, 3 left on base, 1 run 3rd through 9th inning -- 9 at bats, 3 left on base, 3 runs
1 + 3*7 = 22 runs
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extreme


Registered: 04/05/11
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Re: mathematical brain teaser thread [Re: DisoRDeR]
#22037829 - 08/03/15 01:58 PM (8 years, 5 months ago) |
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Under these circumstances, Casey is gonna be moving back to the minors or something after this game, or traded. Maybe even time for him to retire. For SURE isn't leading off anymore! 
DisoRDeR it looks to me like everything checks out. Maybe the only thing that could be wrong that I just noticed is you didn't load the bases in the first inning. That way in the 2nd inning no runs could still be scored yea?
Otherwise your methodology otherwise looks OK to me. My answer would be just adjusted from his to change that 1st/2nd inning, and once that pattern came up where Casey was the last out then the same thing from innings "3-9:"
The 9 at bats, 3 LoB, 3 runs.
So it would be 21 runs if everything else were the same but the 2nd inning would change later innings so it might be like 20 or 19 I might slowly work that out to get an exact answer after a little while. Probably isn't a lot more work lol.
Edit: more jokes about Casey's bad game how about that clutch factor eh? All his teammates are practically hitting off a tee in this game! He sees numerous bases loaded scenarios and always fails most likely the hardest win of his career!
Edited by extreme (08/03/15 02:07 PM)
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DividedQuantum
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Re: mathematical brain teaser thread [Re: DisoRDeR]
#22037972 - 08/03/15 02:41 PM (8 years, 5 months ago) |
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Quote:
DisoRDeR said: Casey always gets out and ends innings, except the first.
1st inning -- 3 at bats, 0 runs 2nd inning -- 7 at bats, 3 left on base, 1 run 3rd through 9th inning -- 9 at bats, 3 left on base, 3 runs
1 + 3*7 = 22 runs
I'm sorry, but that is incorrect.
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