Time Calculation For Lighting Up The Candles

What is the exact situation?

From a moment from first candle is lit, 30 seconds later there would be total 2 candles lit. In next 30 seconds, each of these 2 candle holders will find 1 candle to lit. So there are now 4 candles lit after 60 seconds. In next 30 seconds, these 4 candles would lit another 4 candles making total 8 candles lit. 

In short, for every 30 seconds, the number of candles lit are doubled. So after 7 sets of 30 seconds, 2^7 = 128 candles would be lit. At 8th set of 30 seconds, 256 candles can be lit. But we have only 200 candles. Still 72 of 128 candles would lit another 72 in 8th set of 30 seconds. 

To conclude, 8 X 30 = 240 seconds = 4 minutes required to lit all 200 candles. 

Check Mate in 1 Move

You are playing with white and its your turn. Check mate the opponent in 1 move.

'Here' is that move!

'That' Move to Check Mate The Opponent

First know the current situation on the board!

Just kill the Black Rook by pawn with the move C7 = > B8. Revive KNIGHT at that position to check mate the opponent straightway. Here, opponent can't use bishop at D6 to kill our knight as in that case his king will be in line of attack of our rook at D1.

 

Formations of Special 6-Digit Numbers

How many six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place?

Skip to the count!

Counting The Formations of 6-Digit Special Numbers

You may need to read the question first! 

Let's remind that the numbers can't be repeated. So mathematically there are total 6! (720) unique numbers can be formed.

The number XXXXX1 will be always divisible by 1; so there we have 5! = 120 numbers.

The number XXXXX2 will be always divisible by 2; so there we have 5! = 120 numbers. 

Since sum of all digits is 21 which is divisible by 3; the number XXXXX3 will be always divisible by 3. So we have 5! = 120 more such numbers.

The number XXXXY4 is divisible only when Y = 2 or 6. So in the case we have 2 x 4! = 48 numbers.

The number XXXXX5 will be always divisible by 5; so there we have 5! = 120 numbers.

The number XXXXX6 will be always divisible by 6 (since it is divisible by 2 & 3); so there we have 5! = 120 numbers.

Adding all the above counts - 120 + 120 + 120 + 48 + 120 + 120 = 648.  

So there are 648 six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place.
  

Trip Around The Earth

Professor Fukano plans to circumnavigate the world in his new airplane. But the plane's fuel tank doesn't hold enough for the trip—in fact, it holds only enough for half the trip. But with the help of two identical support planes (which can refuel him in mid-air) piloted by his assistants Fugari and Orokana, the professor thinks he can make it in one trip. But since all three planes have the same problem of limited fuel, how can they work together to achieve the professor's goal without anyone running out of fuel?

1. The professor's plane must make a single continuous trip around the world without landing or turning around.

2. Each plane can travel exactly 1 degree of longitude in 1 minute for every kiloliter of fuel. Each can hold a maximum of 180 kiloliters of fuel.

3. Any plane can refuel any of the others in mid-air by meeting at the same point and instantly transferring any amount of fuel.

4. Fugari and Orokana's planes can turn around instantaneously without burning fuel.

5. Only one airport is available for any of the planes to land, take off, or refuel.

6. All three planes must survive the experiment, and none may run of fuel in mid-air.

'This' is how mission is completed!
  

For The Journey Around The Earth

First read T&Cs of the journey! 

Let's assume that the only airport mentioned is located at the top of the earth. 

Recollect all the data given.
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1. The professor's plane must make a single continuous trip around the world without landing or turning around.

2. Each plane can travel exactly 1 degree of longitude in 1 minute for every kiloliter of fuel. Each can hold a maximum of 180 kiloliters of fuel.

3. Any plane can refuel any of the others in mid-air by meeting at the same point and instantly transferring any amount of fuel.

4. Fugari and Orokana's planes can turn around instantaneously without burning fuel.

5. Only one airport is available for any of the planes to land, take off, or refuel.

6. All three planes must survive the experiment, and none may run of fuel in mid-air.

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As per (2), since each plane travel 1 degree of longitude in 1 minute for every kiloliter of fuel, means that plane need 360 minutes (6 hours) and 360 kiloliters of fuel. But remember plane can hold only 180 kiloliters of fuel.

Let's suppose that, all three planes takes off from airport exactly at 12:00 PM towards the WEST.

We will break this 6 hours journey into 8 parts where each plane travels 45 degree of longitude east or west using 45 kiloliters of fuel.  

START :

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PART 1 :

 Exactly at 12:45 PM, all will be at 45 degree angle with reference to the center of the earth.
At this point, each of them will use 45 kiloliters of fuel. Hence, each will have 135 kiloliters of fuel. Here, Orokana gives away 45 kiloliters to each of Fukano & Fugari. So she is left with the 45 kiloliters which she uses to go back to the starting point.

 
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PART 2 : 

In next 45 minutes, both Fukano and Fugari moves to 90 degree of longitude spending 45 kiloliters of fuel. Now,here both will have 135 kiloliters each in their fuel tank. Here, Fugari refuels Fukano's fuel tank with 45 kiloliters of fuel; leaving 90 kiloliters in own tank for the backward journey towards the starting point.

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PART 3 : 

Fukano travels further, while Fugari is in midway of the backward journey. Again, both spend 45 kiloliters of fuel.

 
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PART 4 :

Exactly at 3:00 PM, Fukano reaches at 180 degree while Fugari reaches back to the starting point. Till then, Orokana refuels her plane & takes off towards EAST. She has to take off as Fukano is left with only 90 kiloliters of fuel by which he could travel half of the rest of journey.

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PART 5 :

In next 45 minutes, Fukano's plane uses 45 kiloliters further while Orokana travels 1/3rd of Fukano's remaining journey in reverse direction so as to meet Fukano in midway. In process, her plane again uses 45 kiloliters of fuel with 135 kiloliters left.

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PART 6 :

At 04:30 PM, Orokana meets Fukano whose plane had no fuel left at the point & refuels his plane with 45 kiloliters of fuel. Remember Orokana's plane consumed 45 kiloliters more till she meets Fukano. Now, since both of them have left only 45 kiloliters in fuel tank, Fugari whose plane standing at airport is refueled at full 180 kiloliters takes of in the direction of EAST.

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PART 7 : 

At 05:15 PM, when fuel tank indicators of both Fukano & Orokana are pointing to 0, Fugari meets them & gives 45 (to Fukano) + 45 (to Orokana)  = 90 out of 135 (180 - 45 used since take off) . Now all are left with 45 kiloliters of fuel & 45 degrees of journey is left.

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PART 8 : 

And this is how, exactly at 06:00 PM all of them reaches back to the starting point safely.

But is this the most efficient way to make trip around the Earth? Certainly not! 

If plane was built with fuel tank of 360 + then the mission wouldn't have required any assistance. Just because of limited fuel tank, 45 + 45 + 90 + 90 + 90 + 90 + 45 + 45 = 270 Kiloliters of fuel burnt to assist Fukano's plane.