crossing the bridge

Four friends A,B,C and D are at one end of bridge and it's night. They have one torch to be used while crossing the bridge. All four have different speeds of walking and they take 1, 2, 5 and 10 minutes respectively to cross the bridge. At a time, at most two people can be crossing the bridge (walking on either of the sides). If say A and D walk together, A has to slow down because they have to go together using just one torch. Let them show how they can cross the bridge given that the torch battery will last only for 17 minutes.

\via{Prashant}, \thanks_for_hint{Prashant}

The starting approach could be trivial in which we try to send 2 of them on other side, let one of them come back with torch and take another guy from first side to second side. Initially we may think that A will be the speediest person to go back to give torch back. With this we will always end up in using torch for at least 3 times for A's back journey.
For example->
A,B,C,D ------------------ None
B,C ---------------------- A,D
A,B,C -------------------- D
....
....
and so on.
But this is not the optimal way because the time when two of them go together is lower bounded by the person who is slower amongst the two. If every time, A takes one of them to the other side, each trip will take 2,5 and 10 minutes respectively. Note that this itself sums up to 17 minutes, but as we saw earlier, at least 3 minutes are required for A's back journey. Thus, this kind of approach will not work.

The quick hint that helps solving this is let C and D go together and keep them steady, as in don't let them come back. If that is to be done, we can send C and D initially itself; but that will require one of them (of course C) to come back which we don't want. So, let's send A and B first and let A_the_speediest come back. Now it can wait and let C and D cross the bridge with the torch. This way C and D cross together and B already on other side can quickly take torch back.

Thus, the scheme will be->
A,B,C,D --------------- None
C,D ------------------- A,B (takes 2 mins)
A,C,D ----------------- B (takes 1 min)
A --------------------- B,C,D (takes 10 mins)
A,B ------------------- C,D (takes 2 mins)
None ------------------ A,B,C,D (takes 2 mins)
Total takes 17 mins.

measure 4L water

You have to measure 4L water with 2 measuring cups, one that can contain 3L water and other that can contain 5L water. Measuring cups obviously don't have per liter markings! :)
\via{Prashant}

My idea was as follows. It is not possible to directly have 4L in 1 measurement as 3 and 5 don't in any way make up 4. So, we have to somehow come up with method which will keep 4L in measuring cups themselves - either separately or combined - say 1+3, 2+2, 0+4 etc. But, it is not possible to have 4L divided up in 2 cups, because in say we want to achieve 1 in 3L and 3 in 5L, we don't have empty cups to measure anything at this point. This indicates that we can only have 0+4 or 4+0 in cups. Obviously, 3L cup cannot contain 4L, so we have to somehow have 4L kept in 5L cup in the end. Starting with this final step backwards, what keeps 4L in 5L cup? One way is to have 5L filled in completely and pour just 1L from it to 3L cup. This means to be able to do this, 3L cup should be already having 2L water in it, so that it can accommodate only 1L more. Getting 2L in 3L cup is fairly easy. So, overall following are the steps- (let A = 3L cup, B = 5L cup)
0. A = B = 0 \\ Both empty initially
1. A = 0, B = 5 \\ Fill in 5Lcup
2. A = 3, B = 2 \\ Pour 3L in A, 2L remains in B
3. A = 0, B = 2 \\ Empty 3L cup
4. A = 2, B = 0 \\ Pour 2L from B to A
5. A = 2, B = 5 \\ Fill B cup again
6. A = 3, B = 4 \\ Pour 1L from B to A, B has 4L remaining

how many zeros in the end of huge number

Note: Came across this Q at TIFR-Ph.D.entrance test.

Let n > 1 be an odd integer. How many zeros are at the end of number S = 99^n+1 (Read as S = 99 raised to the power n and then 1 added to it)

At first glance, this indicates some trick, for two reasons. First, S is going to be huge very fast, even with comparatively small n and second, this was asked in Ph.D test of TIFR! :D

I thought of splitting 99 as 100-1 and getting a generic result form for S.
i.e.

S = 99 ^ n - 1 = P - 1..................... s.t. P = 99 ^ n (n > 1, odd)
so,
P = 99 ^ n = (100 - 1) ^ n = (-1 + 100) ^ n
Using binomial theorem for expansion,
P = \sum_over_k=0_to_k=n {nCk * (-1)^k * 100 ^ (n-k)}
P = nC0 * (-1)^0 * 100 ^ n
   + nC1 * (-1)^1 * 100 ^ (n-1)
   + .....
   + nC{n-1} * (-1) ^ (n-1) * 100 ^ 1  ..................... note n-1 is even
   + nCn * (-1) ^ n * 100 ^ 0              ..................... note n is odd

Solving we get alternate terms positive and negative s.t. first term is positive (due to (-1)^0) and last term negative (due to (-1) ^ n, n being odd)

P = P1 - P2 + P3 - ....... - Pk + 100n - 1

where P1, P2, ..., Pk are terms which are multiples of powers (>=2) of 100.

Thus,
P = a_term_multiple_of_powers>=2_of_hundred +100n - 1

Thus,
S = P + 1 = a_term_multiple_of_powers>=2_of_hundred + 100n - 1 + 1 a_number_multiple_of_powers>=2_of_hundred + 100n = a_number_ending_with_two_zeros.

Thus, S ends with 2 zeros.