Hey everyone! I'm finally back on TH-cam after such a long break. I truly apologize for the delay and appreciate your patience. I’ll share the reason for my absence soon, but I promise to stay consistent with posting new videos. Thank you for sticking around!
Out of the box thinking for 1st question : Although the shortest distance must be the one mentioned, a logical answer would be the one wherein he has to carry the water for the least amount of time, reducing the total effort. Coz his speed would reduce once he has the water
@@LOGICALLYYOURS When the farmer’s speed decreases after picking up water, the path should adjust to minimize total effort, similar to light bending to minimize travel time (Snell's Law). In this case, the farmer’s path would follow an analogy to n1 * sin(i) = n2 * sin(r), where n1 and n2 represent the speeds before and after picking up water."
Final question: Answer is 4x17 and 2x16. First I realised that 40 was useless because it can be made up from 16 and 24 so if it's a solution theres also a solution without it so no point including 40. 39 also seemed suspicious. I tried just using 16s at first but came up with 96. I thought about replacing 1 with a 24 but that was too much. Then I realised I could swap 4 16s with 17 to gain 4 on the total.
The first problem is analogous to the law of reflection of light, represented as n1 * sin(i) = n2 * sin(r), where n1 and n2 are proportional to the speed of the entity (in this case, the farmer) in the respective mediums. Since both mediums are identical (n1 = n2), the equation simplifies to i = r, which corresponds to moving in a straight line if the boundary between the two mediums is crossed.
For the first puzzle, you said the shortest path WITH THE WATER. The distance the farmer travels with an empty bucket is therefore not relevant. He should go to the riverbank at the spot where a line perpendicular to the river intersects the cow shed. This minimizes the distance the farmer travels with water in the bucket.
For the last question - I divided 100 with the shortest number which was 16 - I got 6 with 4 as remainder... So I replaced 4 of those 16s with 17 and that was the answer..
@@3dworld519 idk about algebra, but from a computer science point of view you can setup an equation: 16i + 17j ... + 40n Then since you know that you won't be able to use more than 7 shots (because the smallest point is 16 and 7x16>100) you can use loops to find out what combinations of i,j,k,l,m and n can give you 100. Worst case scenario you will be checking 8⁶ combinations (6 nested loops checking 0-7) which is admittedly alot, but will probly just take a few minutes and you'd get all possible combinations
1. Reflect barn over bank of river, draw straight line between house and barn. 2. Remove area between top and bottom river bank. Draw straight line between villages. 3. There must be a line passing through the exact center of the garden and the exact center of the plot of land. Place a fence along this line, and you will have perfectly distributed this land and garden in half. 4. 2×16+4×17=100.
My answer to Question 1 would be to fill up the bucket on the way back from the cow shed each day. Then the next day when the cow needs water walk in a straight line to the cow shed. Then refill the bucket on the return trip. The overall distance is the same but it allows the cow to get the water most quickly
I took 6 16s(=96), considered the numbers as 0,1,7 and 8(39 and 40 are useless cuz they could be made from other numbers). now I needed to make 4 from (0,1,7,8) with 6 selections. 4*1+2*0=4 . so 4*17+2*16=100:).
I've seen the bridge problem before, but the width of the river wasn't constant, so that threw in a different variable, but the approach was similar. Very nice, and thank you @logicallyyours Amar for these puzzles, I look forward to them every day!
I think there are two solutions for the 2nd question. One is aligning the point A in the path of the straight line and another is aligning B in the path of the straight line in between X and Y Edit: There is also a third solution where you align the midpoint of AB on the straight line in between X and Y
Hey Akash, good to see your comment. Well, if you try from the opposite side (starting from village Y), then also you’ll get the same path as shown in the video, please give it a try. However, your mid point approach won’t give the shortest path, reason being that the villagers need to travel on the land, whereas reaching the midpoint of AB involves travelling on land+water (which is not possible).
Hmmm, for the final puzzle, does the competition have a limit on number of shots? For now I just calculated: 100/16 = 6 remainder 4 which means replacing 4 of 16 with 17 you can get 2x 16 and 4x 17. But I'm unsure if this is the only solution nor am I sure if this solution is the most efficient. What if the rules say less than 5 shots? I mean I can do 16i + 17j + 23k + 24l + 39m + 40n and try all combinations such that i + j + k + l + m + n ≤ 6 but I hope there's a neat trick in the next video or I'll be sad!
A good question indeed, there’s so limit on number of shots… however, we should also prove that there is no other combination. I hope I’ll post a detailed solution in the next video.
@@analoghardwaretops3976 I mean, the maximum shots you can take is 7 cause it's 100/16 rounded up. For problems like these you always estimate the maximum number of attempts you have by dividing the goal with the smallest value available. This way you can estimate the time needed to bruteforce the solution as well as keep track of your own solution. And it just so happens you can get a solution with 6 moves here. As a computer scientist I'd need this value to know I'd need to make a 7 deep tree and use BFS or DFS to find the solution(s).
That won’t give the shortest path because the people won’t be able to walk down on that straight line as it goes into the river till the mid point of the bridge.
As thatreilyrandomguy have said, 39 and 40 are out of the question. There will be 4 numbers left. 16 17 23 24 She can only take 5 or 6 shots. With 4 shots, the maximum score will be 96. With 7 shots, the maximum score will be 112. 16a + 17b + 23c + 24d = 100 ... (1) Let's say she takes 6 shots. a + b + c + d = 6 ... (2) (2) * 16 ---> 16a + 16b + 16c + 16d = 96 ... (3) (1) - (3) ---> b + 7c + 8d = 4 ... (4) c an d can only be 0 so b = 4 and that makes a = 2. 17 * 4 + 16 * 2 = 68 + 32 = 100
I'm guessing the trick to the last puzzle is noticing that you didn't specify the number of arrows we get to use. It's easy to _assume_ it's a standard three arrows you get, but the problem does not specify this, and it's clearly not a standard archery game, anyway, so it's important not to assume things like that. Trying to get to 100 using only three shots is probably what trips people up.
After reading your comment, I must say, I should have mentioned that there is no limit to use arrows. However, the beauty of a puzzle lies in the simplicity of its wording. But overall, I will keep this in mind for future videos.
Is there any systematic way to approach this like using algebra? Even though I found the answer for the last question by random guessing of 2×16 + 4×17
Last question answer could be 2 x 40 and one arrow between 17 and 23 . That would give us a total of 100 with minimum number of arrows. 16x4 and 17x2 participant will lose because she would have won in the first 3 rounds only.
The only solution to #4 is 16+16+17+17+17+17 = 100, and it's findable with a simple process that takes 5 minutes: keep creating new sums of the starting numbers and checking them against the sums you've found so far. The key is to find both an easy way to keep track of what you've checked so far and an easy way to see whether a sum you're considering is already in your list. The easiest way I found was to keep your numbers sorted by their 2nd digit so you know where to write each new sum and you know where to look for 100 minus that sum. ------- Start by writing the five starting numbers in order by their second digit: 23 24 16 17 39 (Note that I don't include 40 since it's makeable from 24+16 and I want as few numbers as possible.) Now sum each of those with each other one, and write the sums into the chart on a new line: 23 24 16 17 39 41 32 62 33 63 34 55 46 56 47 48 40 (The 1st sum I wrote when creating that second line was 23+23 = 46, the 2nd sum was 23+24 = 47, the 3rd was NOT 23+16 = 39, since 39 was already there, etc.) Now check each sum on the current line to see whether 100 minus that sum exists in the chart (look up the subtraction target based on its 2nd digit, so 100-41 = 59, and in the 9s column in the chart, the only number is 39, so there is no 59 yet). When writing new sums, don't include impossibly large sums. E.g., 39+39 = 78, but 78 needs 22 to reach 100, and 22 is smaller than the smallest sum (32) and is not one of the five starting numbers, so 78 is impossible to use. ------- With line 2 not reaching a solution, we create line 3: 23 24 16 17 39 41 32 62 33 63 34 55 46 56 47 48 40 64 65 66 57 67 58 49 50 And when checking it, voila, we see 66 needs 34, which does exist in the chart (line 2). 66 had been created from line 2's 33+33, and 34 was created from line 1's 17+17. 33 was created from line 1's 16+17. So that solution was 33+33+34 which was 16+17 + 16+17 + 17+17 = 100. Other solutions found on that line are 67+33 and 50+50, both of which are from the same combination of 16s and 17s. To verify that these are the only solution, keep creating rows and checking for solutions until all sums attempted are too large (which happens when trying to create an impossible 5th line, and in fact the 4th line includes only two new sums, 83 and 84, both of which have identical solutions to the one already found).
Thanks for taking the time to write out your approach, I am always happy to hear how people solve these puzzles! I will review this process in detail to feature in the solution video where we need to prove why there’s only one combination.
All of the answers that calculate = 100 are wrong… the Question very clearly says closest to 100 and not = 100 … listen to the Question properly … so if you look at natural numbers closest to 100 is 99 or 101… one possible solution is 3x17 and 3x16….the young lady should not have won the first prize
My solution very easy and fun : I cut 6 wooden rod at the size 40mm 39mm etc etc So 6 wooden rod , and make a mark on paper length of 100mm , and try to align my 6 rods for make 100mm on paper Very fun to do and no tricky mentally mathematic calculs 🧮, à children can solve this Game is higher form of research ….
Nice thought,but you need to use one rod more than once ie you may need to use 16 twice,thrice or even more. So you need to have multiple rods of the same length to do this activity successfully
Hey everyone! I'm finally back on TH-cam after such a long break. I truly apologize for the delay and appreciate your patience. I’ll share the reason for my absence soon, but I promise to stay consistent with posting new videos. Thank you for sticking around!
Glad you came back.
Welcome back
All good
It’s so good to see you back I like so much your videos !! 🎉
Nice to see you back! 😎
Out of the box thinking for 1st question :
Although the shortest distance must be the one mentioned, a logical answer would be the one wherein he has to carry the water for the least amount of time, reducing the total effort.
Coz his speed would reduce once he has the water
Brilliant thinking! An important constraint indeed while implementing practical solutions.
@@LOGICALLYYOURS When the farmer’s speed decreases after picking up water, the path should adjust to minimize total effort, similar to light bending to minimize travel time (Snell's Law). In this case, the farmer’s path would follow an analogy to n1 * sin(i) = n2 * sin(r), where n1 and n2 represent the speeds before and after picking up water."
Absolutely right...
@@DeepeshMaan Exactly! It is in the textbook
Then the problem will be extremely easy to solve.
Final question: Answer is 4x17 and 2x16. First I realised that 40 was useless because it can be made up from 16 and 24 so if it's a solution theres also a solution without it so no point including 40. 39 also seemed suspicious. I tried just using 16s at first but came up with 96. I thought about replacing 1 with a 24 but that was too much. Then I realised I could swap 4 16s with 17 to gain 4 on the total.
💯
You’re absolutely right on the spot. 😁
@@LOGICALLYYOURSIs there any systematic way to approach this like using algebra?
Lets write an equation
16a + 17b + 23c + 24d + 39e + 40f = 100. Note that the number of shots is at most 6, since 7x16 would be too much. As such 0
@@FildasFilipi a note: 16a + 17b + 24d + 40f = 100 [17b=68]
The first problem is analogous to the law of reflection of light, represented as n1 * sin(i) = n2 * sin(r), where n1 and n2 are proportional to the speed of the entity (in this case, the farmer) in the respective mediums. Since both mediums are identical (n1 = n2), the equation simplifies to i = r, which corresponds to moving in a straight line if the boundary between the two mediums is crossed.
For the first puzzle, you said the shortest path WITH THE WATER. The distance the farmer travels with an empty bucket is therefore not relevant. He should go to the riverbank at the spot where a line perpendicular to the river intersects the cow shed. This minimizes the distance the farmer travels with water in the bucket.
I agree with you that in practical scenarios, this should be the approach.
Yeah may be not exactly perpendicular mathematically but that's atleast so close to the best approach
17 Pts * 4 = 68
16 Pts * 2 = 32
Total = 100! Simple 😅
For the last question - I divided 100 with the shortest number which was 16 - I got 6 with 4 as remainder...
So I replaced 4 of those 16s with 17 and that was the answer..
Yes that’s perfect explanation
@@LOGICALLYYOURSIs there any systematic way to approach this like using algebra?
@@3dworld519 idk about algebra, but from a computer science point of view you can setup an equation:
16i + 17j ... + 40n
Then since you know that you won't be able to use more than 7 shots (because the smallest point is 16 and 7x16>100) you can use loops to find out what combinations of i,j,k,l,m and n can give you 100.
Worst case scenario you will be checking 8⁶ combinations (6 nested loops checking 0-7) which is admittedly alot, but will probly just take a few minutes and you'd get all possible combinations
I found it ! , she would have targeted on 4 17s and 2 16s
1. Reflect barn over bank of river, draw straight line between house and barn.
2. Remove area between top and bottom river bank. Draw straight line between villages.
3. There must be a line passing through the exact center of the garden and the exact center of the plot of land. Place a fence along this line, and you will have perfectly distributed this land and garden in half.
4. 2×16+4×17=100.
For the archiry problem, the girl will hit point 17 four times and point 16 two times. So total will be 17+17+17+17+16+16=100
Final question answer is 2 times 16 + 4 times 17 = 100.
Is there any systematic way to approach this like using algebra?
My answer to Question 1 would be to fill up the bucket on the way back from the cow shed each day. Then the next day when the cow needs water walk in a straight line to the cow shed. Then refill the bucket on the return trip. The overall distance is the same but it allows the cow to get the water most quickly
Appreciate your efforts💖 Welcome back🎉
Thank you for the warm welcome. 😊
I took 6 16s(=96), considered the numbers as 0,1,7 and 8(39 and 40 are useless cuz they could be made from other numbers). now I needed to make 4 from (0,1,7,8) with 6 selections. 4*1+2*0=4 . so 4*17+2*16=100:).
That’s brilliant!!
3:57 so line XA should be parallel to line BY
Good to see you back. 🎉
I've seen the bridge problem before, but the width of the river wasn't constant, so that threw in a different variable, but the approach was similar. Very nice, and thank you @logicallyyours Amar for these puzzles, I look forward to them every day!
Glad you liked it! The next video will be out soon!
Answer : [17 * 2 (34) + 16 * 1 (16)] * 2 = 50 * 2 = 100
We want your explanation to the archery puzzle.
will be posted in a day.
I think there are two solutions for the 2nd question. One is aligning the point A in the path of the straight line and another is aligning B in the path of the straight line in between X and Y
Edit: There is also a third solution where you align the midpoint of AB on the straight line in between X and Y
Hey Akash, good to see your comment. Well, if you try from the opposite side (starting from village Y), then also you’ll get the same path as shown in the video, please give it a try. However, your mid point approach won’t give the shortest path, reason being that the villagers need to travel on the land, whereas reaching the midpoint of AB involves travelling on land+water (which is not possible).
Walking distance from home to river & river to cowshed , both must be equal..for the shortest distance..
(2 sides of the isosceles triangle)...
First puzzle was law of reflection, second was law of refraction, third was symmetry n fourth was algebra.😂😂😂
:D
3:08, for the second puzzle, position the bridge so that the center of it is on the line between X & Y.
Hmmm, for the final puzzle, does the competition have a limit on number of shots? For now I just calculated:
100/16 = 6 remainder 4 which means replacing 4 of 16 with 17 you can get 2x 16 and 4x 17.
But I'm unsure if this is the only solution nor am I sure if this solution is the most efficient.
What if the rules say less than 5 shots?
I mean I can do 16i + 17j + 23k + 24l + 39m + 40n and try all combinations such that i + j + k + l + m + n ≤ 6 but I hope there's a neat trick in the next video or I'll be sad!
A good question indeed, there’s so limit on number of shots… however, we should also prove that there is no other combination. I hope I’ll post a detailed solution in the next video.
@@LOGICALLYYOURS yay! Looking forward to it
Your reasoning of 6 shots is right..😅
@@analoghardwaretops3976 I mean, the maximum shots you can take is 7 cause it's 100/16 rounded up.
For problems like these you always estimate the maximum number of attempts you have by dividing the goal with the smallest value available. This way you can estimate the time needed to bruteforce the solution as well as keep track of your own solution. And it just so happens you can get a solution with 6 moves here. As a computer scientist I'd need this value to know I'd need to make a 7 deep tree and use BFS or DFS to find the solution(s).
17*4, 16*2
100
Is there any systematic way to approach this like using algebra?
Bro finally came back!
She hit 17 four times and 16 twice. 17x4 + 16x2 = 100.
Last puzzle: 4*17+2*16=100
Got confused for about 30sec. Never met similar logical problem before.
beed waiting for soo long for new riddles!
Thank you for being back 😁
I'm back with more brain-bending challenges 😊
For puzzle 2 draw a straight line between x and y and where that intersects with the middle of the river that's where you put the bridge
That won’t give the shortest path because the people won’t be able to walk down on that straight line as it goes into the river till the mid point of the bridge.
The solution to the last puzzle is Subset sum problem
16+16+17+17+17+17=100
Welcome back🎉
As thatreilyrandomguy have said, 39 and 40 are out of the question.
There will be 4 numbers left.
16 17 23 24
She can only take 5 or 6 shots.
With 4 shots, the maximum score will be 96.
With 7 shots, the maximum score will be 112.
16a + 17b + 23c + 24d = 100 ... (1)
Let's say she takes 6 shots.
a + b + c + d = 6 ... (2)
(2) * 16 ---> 16a + 16b + 16c + 16d = 96 ... (3)
(1) - (3) ---> b + 7c + 8d = 4 ... (4)
c an d can only be 0 so b = 4 and that makes a = 2.
17 * 4 + 16 * 2 = 68 + 32 = 100
I'm guessing the trick to the last puzzle is noticing that you didn't specify the number of arrows we get to use. It's easy to _assume_ it's a standard three arrows you get, but the problem does not specify this, and it's clearly not a standard archery game, anyway, so it's important not to assume things like that. Trying to get to 100 using only three shots is probably what trips people up.
After reading your comment, I must say, I should have mentioned that there is no limit to use arrows. However, the beauty of a puzzle lies in the simplicity of its wording. But overall, I will keep this in mind for future videos.
Is there any systematic way to approach this like using algebra?
Even though I found the answer for the last question by random guessing of 2×16 + 4×17
I came up with a systematic approach which I’d post in the next video.
4*17+2*16
Last question answer could be 2 x 40 and one arrow between 17 and 23 . That would give us a total of 100 with minimum number of arrows.
16x4 and 17x2 participant will lose because she would have won in the first 3 rounds only.
Six arrows will put her at 100 exactly; but I fail to see anything tricky about it.
I tried to give people the answer without actually spoiling it for them.
I’m sure you got it correct 👍
2times(16)+4times(17)
I don't understand why the last question is tricky
I also felt the same, but after asking several candidates I realized that it’s certainly a tricky one!
For 1st question, he would build a cowshed near river
the first two were basic optics questions
Two =16
Two=39
Total =110
arreee!!! ye banda zinda hai!!
Aa gaya hu back 😁 with more puzzles
Are there any requirement of how many shots can be taken?
16 16 17 17 17 17
16+16+17+17+17+17=100;
The only solution to #4 is 16+16+17+17+17+17 = 100, and it's findable with a simple process that takes 5 minutes: keep creating new sums of the starting numbers and checking them against the sums you've found so far. The key is to find both an easy way to keep track of what you've checked so far and an easy way to see whether a sum you're considering is already in your list. The easiest way I found was to keep your numbers sorted by their 2nd digit so you know where to write each new sum and you know where to look for 100 minus that sum.
-------
Start by writing the five starting numbers in order by their second digit:
23 24 16 17 39
(Note that I don't include 40 since it's makeable from 24+16 and I want as few numbers as possible.)
Now sum each of those with each other one, and write the sums into the chart on a new line:
23 24 16 17 39
41 32 62 33 63 34 55 46 56 47 48 40
(The 1st sum I wrote when creating that second line was 23+23 = 46, the 2nd sum was 23+24 = 47, the 3rd was NOT 23+16 = 39, since 39 was already there, etc.)
Now check each sum on the current line to see whether 100 minus that sum exists in the chart (look up the subtraction target based on its 2nd digit, so 100-41 = 59, and in the 9s column in the chart, the only number is 39, so there is no 59 yet).
When writing new sums, don't include impossibly large sums. E.g., 39+39 = 78, but 78 needs 22 to reach 100, and 22 is smaller than the smallest sum (32) and is not one of the five starting numbers, so 78 is impossible to use.
-------
With line 2 not reaching a solution, we create line 3:
23 24 16 17 39
41 32 62 33 63 34 55 46 56 47 48 40
64 65 66 57 67 58 49 50
And when checking it, voila, we see 66 needs 34, which does exist in the chart (line 2). 66 had been created from line 2's 33+33, and 34 was created from line 1's 17+17. 33 was created from line 1's 16+17. So that solution was 33+33+34 which was 16+17 + 16+17 + 17+17 = 100.
Other solutions found on that line are 67+33 and 50+50, both of which are from the same combination of 16s and 17s. To verify that these are the only solution, keep creating rows and checking for solutions until all sums attempted are too large (which happens when trying to create an impossible 5th line, and in fact the 4th line includes only two new sums, 83 and 84, both of which have identical solutions to the one already found).
Thanks for taking the time to write out your approach, I am always happy to hear how people solve these puzzles! I will review this process in detail to feature in the solution video where we need to prove why there’s only one combination.
23 + 29 + 24 + 24
All of the answers that calculate = 100 are wrong… the Question very clearly says closest to 100 and not = 100 … listen to the Question properly … so if you look at natural numbers closest to 100 is 99 or 101… one possible solution is 3x17 and 3x16….the young lady should not have won the first prize
4x17 + 2x16 = 100
blocked your channel because you didn't give the answer to the target riddle.
Gary, the solution will be posted tomorrow for sure!
Clickbait bullshit. No answer given to the actual puzzle.
Hey John, the solution to the final puzzle will be posted tomorrow. Also, I changed the thumbnail of this video.
My solution very easy and fun : I cut 6 wooden rod at the size 40mm 39mm etc etc
So 6 wooden rod , and make a mark on paper length of 100mm , and try to align my 6 rods for make 100mm on paper
Very fun to do and no tricky mentally mathematic calculs 🧮, à children can solve this
Game is higher form of research ….
Nice thought,but you need to use one rod more than once ie you may need to use 16 twice,thrice or even more. So you need to have multiple rods of the same length to do this activity successfully
Thanks @HerveSACUTO for the detailed explanation. I really appreciate your approach.