How to crack problems on Clocks?

There can be different types of clock problems:
A few are mentioned here:
1. Finding angle (very common type and also, an important one)
2. Time taken for hands to travel certain angle
3. Faulty clocks (clocks gaining time or losing time) etc.

For most of these problems, you need to keep angular speed of hands in mind.

Some points to keep in mind:
1. Each hour is of 30 degrees. 3 hours make 90 degrees.
2. Hour hand moves 1/2 degrees per minute (30 degrees in one hour)
3. Minute hand moves 6 degrees per minute (360 degrees in one hour)
4. From one and two, minute hand is 12 times faster than hour hand.
5. A faulty clock which is gaining time, moves faster than expected.
A faulty clock which is losing time, moves slower than expected.
When a clock is faulty, both hands are faulty.

Today's 5 questions

1.     Find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

2.     5789^(1!+2!+3!-------+1000!)has the unit digit of ?

3.     Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4 < a < 8, how many values of N are possible?
a. 1 
b. 2 
c. 3 
d. 4 
e. more than 4

Innovative shortcuts for finding Compound Interest without calculation...

We all know the traditional formula to compute compound interest.

CI = P*(1+R/100)^N - P

This calculation gets very tedious when N>2 (more than 2 years). The method suggested below is a very simple way to get CI/Amount after 'N' years.

You need to recall the old Pascal's Triangle in following way:

Code:

 

Number of Years (N)                     

  -------------------

        1                        1

        2                    1   2   1

        3                  1   3   3   1

        4                1   4   6   4   1

        .              1 .... .... ... ...   1

Example: P = 1000, R=10 %, and N=3 years. What is CI & Amount?

Step 1: 10% of 1000 = 100, Again 10% of 100 = 10 and 10% of 10 = 1

We did this three times becoz N=3.

Step 2:

Now Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331/-

The co-efficients - 1,3,3,1 are lifted from the pascal's triangle above.

Playing with Numbers - Part 1

1) 2^2n-1 is always divisible by 3 
Proof: 2^2n-1 = (3-1)^2n -1 
                      = 3M +1 -1 
                      = 3M, thus divisible by 3 

2) What is the sum of the divisors of 2^5.3^7.5^3.7^2? 
Ans : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6

Rules: if a number 'n' is represented as 
a^x * b^y * c^z .... 
where, {a,b,c,.. } are prime numbers then 

Remember:

(a) the total number of factors is (x+1)(y+1)(z+1) .... (b) the total number of relatively prime numbers less than the number is n * (1-1/a) * (1-1/b) * (1-1/c).... (c) the sum of relatively prime numbers less than the number is n/2 * n * (1-1/a) * (1-1/b) * (1-1/c).... (d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x*y*...)