Solving it mathematically..

I              x              I   y/4    I               3y/4                  I
I———————I———–I——————————I
A                           B             C                                          D

A = is the train’s current position,
B = Start of tunnel
C = man’s current position
D = end of tunnel

AB = x miles
BD = y miles; so BC = y/4 and CD = 3y/4

Man’s speed is 20mph. Let S be the train’s speed.

Let the man take Tb time to get back to the start of the tunnel
So, 20mph = y/4 miles / Tb or Tb = y/4 miles / 20 mph
In this time, the train will also just make it to the entrance of the tunnel;
So Tb = x / S

So we can say that Tb = y/80 = x/S or Sy = 80x—Eq1

If the man runs in the other direction, toward the end of the tunnel, then, let him take time Td
Td = 3y/4 miles / 20
By this time the train would have reached the end of the tunnel as well.
So Td = (x+y)/S
So Td = 3y/80 = (x+y)/S or S3y = 80x + 80y—Eq2

Substituting for 80x from Eq1, Eq2 now becomes S3y = Sy + 80y or S3y = y(S+80) or S3 = S + 80 or S2 = 80 or S = 40.

So the speed of the train is 40mph.

Good one by Yodha. I came to the same answer, but without considering the Japanese flag specifically.

The explanations given by everyone else had no “public” element to it. As in, they were all doing something in private and what they were doing was not contingent on the state of affairs existing prior to the theft. They weren’t changing a condition that existed prior to the theft. So unless the captain was with them personally,  there is no way to contradict their statements.

The housekeeper was the only one who said that he changed a flag. Now a flag is something that is out there. Can be seen by anyone. And also, he “changed” the state of the flag. So if the captain had known that the flag was right in the first place, then he could have picked up on this.

This is not as elegant a solution as the one given by Yodha, but it does cover the eventuality of the flag being from any country or even Jack Sparrow’s 🙂

Cheers

obama or the snowman 🙂

There are multiple solutions ofcourse, find below the way I thought through this;

Total milk is 3321, so each son should get 369.

If you write the full sequence (1 to 81) in a straight line and then add the extreme numbers in pairs (1+81, 2+80, 3+79 and so on), each pair will add up to 82. 4 such pairs will add upto 328. So lets give 4 such pairs to each son, so 8 cows to each son. Each of them will have 328 with them. The order looks like below;

Son 1 – 01, 02, 03, 04 and 81, 80, 79, 78 – the first 4 and the last 4 cows
Son 2 – 05, 06, 07, 08 and 77, 76, 75, 74 – the next 4 cows from either end
Son 3 – 09, 10, 11, 12 and 73, 72, 71, 70 – and so on..
Son 4 – 13, 14, 15, 16 and 69, 68, 67, 66
Son 5 – 17, 18, 19, 20 and 65, 64, 63, 62
Son 6 – 21, 22, 23, 24 and 61, 60, 59, 58
Son 7 – 25, 26, 27, 28 and 57, 56, 55, 54
Son 8 – 29, 30, 31, 32 and 53, 52, 51, 50
Son 9 – 33, 34, 35, 36 and 49, 48, 47, 46

Now there are 9 cows remaining 37 to 45.

To get to 369, each of them need 41 more. 41 is the 5th in sequence. So lets give 41 to the 5th son. Now he has 328+41 = 369. He is sorted.

Now consider the remaining sons in pairs (Son 1 and Son 2), (Son 3 and Son 4), (Son 6 and Son 7), (Son 8 and Son 9)
If you give 37 to son 1, then he is 4 short. So swap a pair of cows between the two sons that has a difference of 4 (basically any cow), Say they swap their first cows

So at this stage,
Son 1 – 05, 02, 03, 04 and 81, 80, 79, 78 = 332+37 = 369
Son 2 – 01, 06, 07, 08 and 77, 76, 75, 74 = 324. So he needs a cow that covers for the 41 he was short in the first place and the 4 he gave to Son 1  – so the only option is 45. So this pair is sorted.

Take the next pair. Each of them are 41 short, but we only have cows which are +/-3, +/-2 and +/-1 away from 41. For this pair, lets do with +/-3. So give 38 to Son 3. He will now be 3 short. So swap a cow with a difference of 3. There are 6 possibilities. Lets go with 10 from Son 1 is swapped with 13 from Son 2.
Son 3 – 09, 13, 11, 12 and 73, 72, 71, 70 = 331+38 = 369
Son 4 – 10, 14, 15, 16 and 69, 68, 67, 66 = 325+44 = 369

For the next pair, they are again 41 short, but we only have cows which are +/-2 and +/-1 away from 41. If we do +/-2 for this pair, then give 39 to Son 6. He will now be 2 short. Swap a cow with Son 7 with a difference of 2. So Son 6 will now gain 2 and be equal to 369. Son 7 will now need to make up for the 41 he needed anyway and the 2 he lost to Son 6. The only option is 43.
Son 6 – 21, 22, 25, 24 and 61, 60, 59, 58 = 330+39 = 369
Son 7 – 23, 26, 27, 28 and 57, 56, 55, 54 = 326+43 = 369

For the final pair, we only have cows 40 and 42 remaining (+/-1 away from 41). Give 40 to Son 8. He will now be 1 short. Swap a cow with Son 9 with a difference of 1. So Son 8 will now gain 1 and be equal to 369. Son 9 will now need to make up for the 41 he needed anyway and the 1 he lost to Son 8. The only option is 42.
Son 8 – 29, 30, 31, 33 and 53, 52, 51, 50 = 329+40 = 369
Son 9 – 32, 34, 35, 36 and 49, 48, 47, 46 = 327+42 = 369

As you can see, in this method alone, there are many combinations possible.

Hope you had the patience to read through this 🙂

Cheers