The legendary king Midas possessed a huge amount of gold. He hid this treasure carefully: in a build - A fair division of Midas' coins is indeed possible. Let the number of rooms be N. This means that per room there are N boxes with N coins each. In total there are N×N×N = N3 coins. One box with N coins goes to the barber. For the six brothers, N3 - N coins remain. We can write this as: N(N2 - l), or: N(N - 1)(N + l). This last expression is divisible by 6 in all cases, since a number is divisible by 6 when it is both divisible by 3 and even. This is indeed the case here: whatever N may be, the expression N(N - 1)(N + l) always contains three successive numbers. One of those is always divisible by 3, and at least one of the others is even. This even holds when N=1; in that case all the brothers get nothing, which is also a fair division! . - Logical & Aptitude Interview Questions and Answers

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Question: The legendary king Midas possessed a huge amount of gold. He hid this treasure carefully: in a building consisting of a number of rooms. In each room there were a number of boxes; this number was equal to the number of rooms in the building. Each box contained a number of golden coins that equaled the number of boxes per room. When the king died, one box was given to the royal barber. The remainder of the coins had to be divided fairly between his six sons. Is a fair division possible in all situations?

Answer: A fair division of Midas' coins is indeed possible. Let the number of rooms be N. This means that per room there are N boxes with N coins each. In total there are N×N×N = N3 coins. One box with N coins goes to the barber. For the six brothers, N3 - N coins remain. We can write this as: N(N2 - l), or: N(N - 1)(N + l). This last expression is divisible by 6 in all cases, since a number is divisible by 6 when it is both divisible by 3 and even. This is indeed the case here: whatever N may be, the expression N(N - 1)(N + l) always contains three successive numbers. One of those is always divisible by 3, and at least one of the others is even. This even holds when N=1; in that case all the brothers get nothing, which is also a fair division! .

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Question: The legendary king Midas possessed a huge amount of gold. He hid this treasure carefully: in a building consisting of a number of rooms. In each room there were a number of boxes; this number was equal to the number of rooms in the building. Each box contained a number of golden coins that equaled the number of boxes per room. When the king died, one box was given to the royal barber. The remainder of the coins had to be divided fairly between his six sons. Is a fair division possible in all situations?

Answer:

A fair division of Midas' coins is indeed possible. Let the number of rooms be N. This means that per room there are N boxes with N coins each. In total there are N×N×N = N3 coins. One box with N coins goes to the barber. For the six brothers, N3 - N coins remain. We can write this as: N(N2 - l), or: N(N - 1)(N + l). This last expression is divisible by 6 in all cases, since a number is divisible by 6 when it is both divisible by 3 and even. This is indeed the case here: whatever N may be, the expression N(N - 1)(N + l) always contains three successive numbers. One of those is always divisible by 3, and at least one of the others is even. This even holds when N=1; in that case all the brothers get nothing, which is also a fair division! . Source: CoolInterview.com

1. At 6’o a clock ticks 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12’o clock

Answer: 60 Seconds

2. A hotel has 10 storey. Which floor is above the floor below the floor, below the floor above the floor, below the floor above the fifth.

Answer: 5th floor.

3. Two trains starting at same time, one from Bangalore to Mysore and other in opposite direction arrive at their destination 1 hr and 4 hours respectively after passing each other. How much faster is one train from other?

Answer: 4 times faster than the other train.

4. A man collects cigarette stubs and makes one full cigarette with every 8 stubs. If he gets 64 stubs how many full cigarettes can he smoke?

Answer: 9 cigarettes.

5. There is one room with 3-bulbs inside and corresponding switches are outside the room. You make any combination of three switches and enter room only once. How do you find out the respective switches for these three bulbs.

Answer: I will switch on the first switch and wait for 5 minutes and then i will turn it off. Then switch on the second switch and then go to the room.
If the bulb is on then its the second switch.
If the bulb is off and cool then its the third switch.
If the bulb is off and hot (as had switched on the first switch for 5 min) then its the first switch
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Answered by: Kiran Kurhade | Date: 4/1/2008 |

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