In Island puzzle only 4 trips are needed. 1St trip (5,3,1)→and come back with 1 2nd trip (6,2,1)→ and come back with 1 3rd trip (8,7,1)→ and come back with 3 4th trip (9,4,3)→ completed
3Ants and Triangle- Collision doesn’t happen only in two cases - All ants going in (a)clockwise (b)anti-clockwise Every ant has two choices and there are total 2^3 possibilities = 8. Out of 2^3 possibilities, only 2 don’t cause collision. probability of ants doesn't collide is 2/8 =1/4 =0.25.
56:05 given batteries b1, ..., b8 of which 4 work, you need only try at most 6 pairs of batteries - not 7 pairs! - in order to find a working pair: 1. if none of the pairs {b1,b2}, {b2,b3}, {b3,b1} work, then you know that at least 2 of {b1, b2, b3} are faulty. 2. likewise, if none of the pairs {b4,b5}, {b5,b6}, {b6,b4} work, then you know that at least 2 of {b4, b5, b6} are faulty. 3. therefore you either found a working pair among {b1, ..., b6} or else you know that all 4 of the faulty batteries are in {b1, ...,b6} - in which case the remaining b7, b8 must both work. thus you only need at most 6 tests to find a working pair of batteries.
The question asks how many pairs you need to TEST. you don't need to TEST the 7th pair {b7,b8}. Either you will have found a pair in 6 or fewer tests, or else you'll be GUARANTEED that {b7,b8} will get the torch on. Please give a close read to how the question is posed and to my solution.
#10 ... n² = (n²-1)+1 = (n+1)(n-1)+1, so if you do not have an even x even grid you can convert it to one with 1 remainder. By subtracting one number from each row of (n-1) you have (n-1)(n+1) - (n-1) + 1 = (n-1) ((n+1) -1) + 1 = (n-1)(n-1+1) + 1 = (n-1)(n) + 1. So to add it back (n-1)(n) + 1 + (n-1). So it equals in this case (9)(9-1) + R + column of 8 numbers converted to a row = 9x9. This is just the math way of saying you can take one number from each row, and place it in a new row underneath (rotate 90 degrees) with the remainder and have an n x n matrix or grid. Drop down the remainder, n² (81), to the last row and slide the right hand section to fill the spaces. The first number of the nth row is the remainder. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 ← take this 10x8 + the Remainder 81 & convert to a 9x9 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 1 2 3 4 5 6 7 8 ↓ 10 ← 11 12 13 14 15 16 17 ↓ 19 20 ← 21 22 23 24 25 26 ↓ 28 29 30 ← 31 32 33 34 35 ↓ 37 38 39 40 ← 41 42 43 44 ↓ 46 47 48 49 50 ← 51 52 53 ↓ 55 56 57 58 59 60 ← 61 62 ↓ 64 65 66 67 68 69 70 ← 71 ↓ 73 74 75 76 77 78 79 80 ← 81 72 63 54 45 36 27 18 9 ← new row. 1 2 3 4 5 6 7 8 10 11 12 13 14 15 16 17 19 20 21 22 23 24 25 26 28 29 30 31 32 33 34 35 37 38 39 40 41 42 43 44 46 47 48 49 50 51 52 53 55 56 57 58 59 60 61 62 64 65 66 67 68 69 70 71 73 74 75 76 77 78 79 80 81 72 63 54 45 36 27 18 9 9 columns of 9 that add to 369. I hope this posts correctly.
In the 50th Puzzle (scenario 1), I have a much more straightforward and quick solution. 👇 First, fill the 5-litre bucket entirely and pour 3 litres of water into a 3-litre bucket. Now, shift that remaining 2 litre of water into 8 Litre bucket. Do this step again, and you will get the 2+2 = 4 Litre water in the bucket of 8 Litre capacity 😌
In the gold bar, king and worker question-- What if the worker sells the 1/7th bar which he gets after the day 1 of work and purchases something? The question shouldv'e mentioned that the worker does not uses the gold bar for till the end of the 7th day to make any purchase.
About the thief in the cave one, is there a reason not to catch the guy in a sandwich by going C1 C13, then C2 C12, then C3 C11, etc? That would make it 7 days instead of 12.
thankyou for this helpful video !!!!!!!!! In a bee travelling between trains why we didn't took relative speed of bee to the trains ? won't it affect ?
#1 ... why do you limit to 8 balls ? it works with 9 too (it's a divide by 3 problem, each use selects one of 3 groups ... with 3 uses you can find among 27 balls and in more general case with n uses among 3^n balls)
13 caves and thief puzzle answer is not optimal. Optimal answer is 8 I think. Instead os always checking 13th cave, they can go clockwise and anticlockwise at the se time. i.e. c12 & c13 on 1st day, c11 & c1 on 2nd day, c10 and c2 on the 3d day, etc. Hope this makes sense
Same, he left out a lot of important information in the question. He also did not say whether balls could be removed from boxes, or how much could be weighed at one time on the scale.
The thief and 13 caves: the answer is 7 Day 1 check c1 and c13 Day 2 check c2 and c12 Day 3 check c3 and c11 Day 4 check c4 and c10 Day 5 check c5 and c9 Day 6 check c6 and c8 Day 7 catch thief as he exits cave 7.
I suspect quite a bunch of these puzzles have a deficiency. E.g. puzzle #7. The solution is to take 1 ball from box 1, 2 balls from box 2 etc. But what if there aren't enough balls in the respective box? E.g. box 10 has less than 10 balls, box 9 has less than 9 balls and so on.
For #24, I agree with Joe. Cops pick two adjacent caves and each day methodically check the next two caves moving around the circle in both directions. The MAXIMUM the thief can hold out is 7 days. There must be some confusion in the way the puzzle is presented. I don't understand what is meant by the third note.
Kevin, I agree with you and Joe, I had the same thought as you (or a similar one at least). Simply Logical has done us a service by assembling so mant puzzles and demonstrating so many solutions. That is a lot of work. It appears that others have had the same thought about the caves. Caroline and Tom made similar points. The police could close in on the thief twice as quickly by moving BOTH cave checks. The police could start out by looking in adjacent caves and then move round the circle in opposte directions by one cave per day until they eventually meet up again. The thief would be caught in between them in an ever diminishing safe zone. Perhaps this is what the third note was about. All the same, I do not see what is different about one policeman and the other.
The puzzle to me about #24 is: What is the miscommunication about how the puzzle is presented? Why does his 12 day solution involve 1 cop staying in one cave? The solution I presented offers a maximum of 7 days, not a minimum. More likely less than 7 days. Looking up the problem shows many people concluding 12 days, more than those who say 7. Some say the thief can evade the cops forever. Something is missing in the presentation of this puzzle.
@@kevinevans3021 I see no reason why one police check should remain stationary while the other moves. Either the thief can pass through the police check--in contradiction to note three--or s/he cannot. Neither the direction of travel nor the movement of the police toward the thief should matter. Your solution of 7 days maximum stands if the thief cannot move into a cave checked by the police the day before.
In last puzzle (i.e the bucket one) we can also follow this approach first fill the 3l bucket put it into 8l bucket, then again fill 3l bucket and put it into 8l bucket, now 8l bucket has 6l of water, now put 5l of water from 8l bucket in 5l bucket this will give us only 1L of water left in 8l bucket now again fill 3l bucket and put it in 8L bucket
Or we can fill 3l bucket, put it into 5l bucket, fill 3l bucket again, put it into 5l bucket again, put the water from 5l bucket to 8l bucket, put 1l from 3l bucket to 5l bucket, fill 3l bucket again and put the 3l to 5l bucket and now we have 4l inside the 5l bucket.
in the defective box one puzzle, i feel we cant assume there exist enough balls in each box to carry out that solution? There could be 2 balls in each box or 3. it should be specified for completeness that there are 10 balls in each box
1:35:20 this is wrong, u can catc hhim in 7 days, also with the same explanation provided afterwards (in regards to the 3rd bulletpoint), because the cop in cave 12 can also move clockwise, because if he would switch with the thief, he would catch him. so there is no point for the other cop to stay stationary in cave 13
Regarding #4 - Ants on a triangle, there seems a simpler solution. Ant 1 moves in either direction. Chance Ant 2 won’t collide with Ant 1 = .5. Chance Ant 3 won’t collide with Ant 2 = .5. .5 x .5 = .25.
That kingdom sounds like Queensland; what a stupid kingdom! The woman could have turned back at the 4 minute mark so that she appeared to be 5 eighths of the way across coming into the kingdom having been seemingly been walking in the same direction for 5 minutes.
Last puzzle solution is wrong. It can be done in 4 steps only if we use 5L bucket first. 1. Fill the 5L bucket and emty it in 8L bucket 🪣 2. Again fill the 5L bucket and trasfer 3 L water in 8L bucket as it already has 5L water already in it. 3. 2L Remaining water in 5L bucket. Emapty it in 3L bucket. 4. Again fill 5L bucket and trasfer 1 L to 3 L 🪣 and thats it. You have 4 litter water
Puzzle 24. Thief and 13 caves: if instead of checking cave 13 each day, they check the next cave instead (ie Day1 the police check 12 & 13. Day 2 Police check 11 & 1... then on the 6th day they will check caves 7 & 5. Day 7 they have their thief. The answer is 7, not 12.
The slave and poison barrel one is wrong on so many levels ... THE worst being you don't have to test one of the barrels ... if no-one dies, then it's the poison barrel.
For defective box puzzle, I see another solution too, according to the problem we've to use the weighing machine only once, but no rule to check the weight with gradual increase right? So instead of selecting in a Arithmatic progression, just select 1 ball from each box, We know ball weighs 10 gm but defective ball, so 10 boxes x 10 balls = should weight 100 gm. So if i gradually pick and add balls in the weighing machine I would be able to notice which box has the defective contains since the weight deviates from multiples of 10. Correct me if I'm wrong
basically I could understand only those puzzles which had visible text on the screen. Because of accent I couldn't understand almost half of other ones correctly
You can cut the gold bar into 4 pieces using a crossing cut (1/8, 1/8, 2/8. 4/8). 1/8 day 1 = 1/8, 1/8+1/8 day 2 = 2/8, trade a 2/8 for a 1/8 day 3 = 3/8, trade 4/8 for 1/8 & 2/8 day 4 = 4/8, 1/8 day 5 = 5/8, 1/8 day 6 = 6/8, trade 2/8 for 1/8 day 7 = 7/8. You keep 1/8 for yourself.
So I submit that the video's example is wrong, in that a King doesn't become a King his way. A King doesn't give away anymore than he has to, there is no reason to give away the entire bar.
The 13 thieves and cave puzzle makes no sense. If the cops start from two caves adjacent, c1 and c13, they can check c2 and c12 the next day, then c3 and c11 and so on. The thief wil not be able to pass the cops, and therefore will be caught in 7 days maximum in the middle cave c7. And that doesn't take into account that the cops can literally catch him on day 1 if they're lucky, which makes the minimum... 1.
#1 - I did not know that we were supposed to assume that all of the lighter balls were of equal weight. #2 - We were not informed that the faster people could be allowed to slow down.
@@elitemaths4994Yeah. I don't know man. I was just as sideways on number 2. I was drinkin' a lot back then or something, so... My apologies, I think?
In torch and batteries puzzle..... If we divide in 4 pairs of 2 batteries in each pair.....b1b2---b3b4---b5b6---b7b8 and test one by one. Worst case you need 4 attempts.That means each pair has a defective battery. Then take any two pair suppose b1b2---b3b4.......and check by trying combinations b1b3 and b2b4.....i.e. 2 trials. So total trials =4+2=6 (less than 7) Correct me if I'm missing any case.
