# Question 16

## Question 16

Swimming deeper, a huge sewer comes in the line of sight of Mario. He dives inside it and the sewer closes. Swimming lower and lower, Mario finds himself inside Bowser’s Castle, where he can finally rescue Princess Peach. Out of the water, he finds himself staring at a huge door, with the heading ‘Bowser Inside’!

Question 16 :

Mario moves on to find a way to get through the main gates of Bowser’s chamber. He suddenly finds falling through a trapdoor and lands on something hard. The Devil Ghost Banshee , comes out and says, “This room is a standard 8×8 chessboard. Each of its 64 square is assigned a weight. These weights are assigned in such a manner that weight of a square is an average of the weight of the square that it is surrounded by.” Mario now has to determine the weights of all 64 squares and for that Banshee tells him the weights of X squares. What could be the minimum value of X with which Mario can deduce the weights of every other square?

a) 8 b) 32 c) 56 d) None of these

Question 16 :

Mario moves on to find a way to get through the main gates of Bowser’s chamber. He suddenly finds falling through a trapdoor and lands on something hard. The Devil Ghost Banshee , comes out and says, “This room is a standard 8×8 chessboard. Each of its 64 square is assigned a weight. These weights are assigned in such a manner that weight of a square is an average of the weight of the square that it is surrounded by.” Mario now has to determine the weights of all 64 squares and for that Banshee tells him the weights of X squares. What could be the minimum value of X with which Mario can deduce the weights of every other square?

a) 8 b) 32 c) 56 d) None of these

**TECHNOTHLON**- Admin
- Posts : 39

Join date : 2016-07-19

## Re: Question 16

Answer should be (d) None.

Because all the squares actually have the same weight which we get from the data.

Hence minimum value is 1.

Because all the squares actually have the same weight which we get from the data.

Hence minimum value is 1.

**Saurav.S**- Posts : 6

Join date : 2016-07-21

## Re: Question 16

Well,

If you consider the square with the highest weight.The average of all weights surrounding it should be equal to its weight.Hence all the neighbouring squares should have value equal to the weight, since average will always be equal or less than one of the squares. Hence eventually we get all squares should have equal weight assigned to it.

If you consider the square with the highest weight.The average of all weights surrounding it should be equal to its weight.Hence all the neighbouring squares should have value equal to the weight, since average will always be equal or less than one of the squares. Hence eventually we get all squares should have equal weight assigned to it.

**Saurav.S**- Posts : 6

Join date : 2016-07-21

## Correction

I dont feel you can ask things like the heviest one. i feel 32 will be required

**Aryan Mundada**- Guest

## Re: Question 16

I think 8 though what I marked I don't remember

**Gitanjit**- Posts : 20

Join date : 2016-07-27

## Re: Question 16

I think 8 though what I marked I don't remember

**Gitanjit**- Posts : 20

Join date : 2016-07-27

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