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*...)

3) what is the highest power of 10 in 203!

Ans : express 10 as product of primes: 10 = 2*5

divide 203 with 2 and 5 individually

203/2 = 101

101/2 = 50

50/2 = 25

25/2 = 12

12/2 = 6

6/2 = 3

3/2 = 1

thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198

divide 203 with 5

203/5 = 40

40/5 = 8

8/5 = 1

thus power of 5 in 203! is, 49

So the power of 10 in 203! is 49 (we select 49 because 49<198)

N.B: The main reason goes in this way - we need to find the power of 10 that means how many zeroes are there at the end and 0 is equivalent to 10 and 10 is equivalent to (2*5) - so 198 two's and 49 five's can together make only 49 ten's. So the power of 10 in 203! is 49.

4) How many zeroes are there at the end of 203!

 Ans: Same as above question. The procedure is also the same.