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.

Prime number pairs summing to an odd number

While reading a book (details below), I came across a very nice problem as follows.

Find all pairs of prime numbers that sum up to 999.

As the book says, it involves thinking before actually attacking the problem. Few observations:

• Sum given is odd (999).
• To get sum odd, exactly one of the two numbers have to be even.
• Asked are the pairs of prime numbers.
• Meaning, we want intersection of prime and even which is just the number 2.

So, one of the numbers has to be 2; and thus the other of course is 997. And, that's the only pair of prime numbers which sums up to  999.

Cool indeed! :)

Book details:
Title:  math Charmers - Tantalizing Tidbits for the mind
Author: alfred s. posamentier
Publisher: University Press