Solve it, If you can
Answers
Answer:
1.First trip to the other side: man + goat
Second trip: man + lion (when getting to the other side man loads goat again and goes back with the goat)
Third trip: man drops off goat and takes grass ... then goes back)
Fourth trip: drops off grass where the lion already is
Fifth trip: goes back and picks up goat for the second time
Sixth trip and last trip: goes back with the goat to where the lion and the grass already are... the end.
2.Idk...
3.The largest number of pigeonholes, n to hold the 100 pigeons would be obtained by having the least number of pigeons in each hole, and not having the same amount of pigeons in any two holes.
Start by placing 1 pigeon in the 1st hole, 2 in the second hole, 3 in the third hole, etc., until the total number of pigeons in the all occupied (n) pigeonholes is 100.
The number of pigeons in the n holes will then be given by the formula for the sum of the first n natural numbers which is n*(n+1)/2.For n = 13, we get 13*(14)/2 = 91 pigeons. Therefore, we need to accommodate another (100 - 91) 9 pigeons.
Placing the last 9 pigeons in a 14th pigeonhole would violate the requirements, as there would be two holes with 9 pigeons.
If we use n = 14, we would need 14*15/2 = 105 pigeons for this scheme to work.Placing an extra pigeon in each of the last 9 of the 13 holes would be one way to solve the problem.
So, pigeonhole 5 will have 6 pigeons, pigeonhole 6 will have 7 pigeons, etc., etc.
Check : 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 +14 = 100.
Therefore, the maximum number of pigeonholes required is 13.
4.This question is Incomplete.
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HOPE THIS HELPS