5839
points
Questions
3085
Answers
236
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I BET you cant
I won the bet, because NO DIFFERENCE 😛
This answer accepted by SherlockHolmes. on 29th March 2020 Earned 20 points.
- 5227 views
- 1 answers
- 0 votes
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- 10942 views
- 1 answers
- 1 votes
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1. Liquid state
2. It caused a revolution
3. It will Sink
4. So easy, sleeps at night.
5. You will never find an elephant with one hand.
6. Concrete floors wont crack.
7. No time at all it is already built.
8. Very large hands.
9. The other half.- 7434 views
- 1 answers
- 0 votes
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I Got the VASE a bit later, until i thought its just two baby facing each other
- 3952 views
- 2 answers
- 0 votes
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Answer : 366329
PROCESS-
a+b+c
First 2 digits = a*b
Next 2 digits = c*a
Last 2 digits = Reverse of (((a*b)+(a*c))-c)
Solution-3+5+6
First 2 digits = a*b = 15
Next 2 digits = c*a = 18
Last 2 digits = Reverse of (((a*b)+(a*c))-c) = Reverse(15+18-6) = Reverse(27) = 72
Final value 151872Similarly,
For Given Question –9+4+7
First 2 digit = 36
Next 2 Digit = 63
Last Digits = 29So, Solution – 366329
This answer accepted by SherlockHolmes. on 24th August 2015 Earned 20 points.
- 3989 views
- 1 answers
- 1 votes
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I know you don’t have Bad Eyes!!! just try again
Answer- BAD EYES 😛
Trick-
Trick is in the visibility of eyes, close your eyes 90% and then see.
You will see BAD EYES now.- 5864 views
- 1 answers
- 0 votes
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ALGORITHM:
Start with the initial coordinates of the knight.
Make all possible moves for this position and multiply these moves with their probability, for each move recursively call the function continue this process till the terminating condition is met. Terminating condition is if the knight is outside the chess board, in this case return 0, or the desired number of moves is exhausted, in this case return 1.As we can see that the current state of the recursion is dependent only on the current coordinates and number of steps done so far. Therefore we can memorize this information in a tabular form.
For better understanding and Computer program of the algo – crazyforcode.com
- 7339 views
- 2 answers
- 1 votes
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First thought which comes to our mind is to use binary search, we first drop Egg#1 from 50th floor, if it does not break, then try the middle of second half, if breaks then we have to try each floor in first half. But this will give worst case number of drops as 50(if it brakes on 50th floor, then we have to try from 1 to 49 floors sequentially).
Second thought is to try xth floor then 2xth floor till 100th, in this case worst case time will be (100/x)+(x-1). worst case will be when Egg #1 breaks at 100th floor then we have to try Egg #2 from (100-x)th to 99th floors. In (100/x) + (x-1) equation, with increase in x, 100/x decreases while (x-1) increases, thus we can minimize it when 100/x = (x-1), this gives x ~10, which gives worst case number as 19 drops.
But increasing the floor every time by x is not a very nice idea, as with each new increase in Egg #1 drop, we should decrease Egg #2 drops to minimize worst case number. so if we drop Egg #1 from xth floor initially, then in next turn we should try x + (x-1)th floor(to keep the worst case number same).
Thus we can say X + (x-1) + (x-2)…1 = 100
X(X+1)/2 = 100 => X=14.So we should drop Egg #1 from 14th, then 27th, then 39th and so on.
- 174246 views
- 15 answers
- 2 votes
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Three Ducks
They’re all in a row, so ducks 1 and 2 are in front of duck 3, ducks 2 and 3 are behind duck 1 and duck 2 is in the middle.
- 24671 views
- 3 answers
- 0 votes
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- 3093 views
- 1 answers
- 0 votes