Sunday, September 30, 2012

How will you represent (3AC)14 in base 7 numeral system.

It can be done using this formula!
If a no X is represented in base 14 as
X = an a(n-1) a(n-2) .... a(0)
then in base 7 we can write it as
X=.....rqp
where
p=(2^0)a(0)%7;
q=((2^1)a(1) + p/7)%7
r=((2^2)a(2) + q/7)%7
..........
nth term=((2^n)a(n) + (n-1)th term/7)%7
(will go further because a no. in base 14 will require more digits in base 7).
The logic is simple, just based on properties of bases, and taking into account the fact that 7 is half of 14. Else it would have been a tedious task.
Eg. here it is given 3AC.
C =12;
so last digit is (2^0 * 12)%7 = 5
A=10
next digit is (2^1 * 10 + 12/7)%7 = (20+1)%7=21%7=0
next is 3;
next digit is (2^2 * 3 + 21/7)%7 = (12+3)%7=15%7=1
next is nothing(0);
next digit is (2^3 * 0 + 15/7)%7 = (0+2)%7=2%7=2
Hence, in base 7 number will be, 2105.

Now try how you can perform any base to any other base conversion ?

You are given n dice. Each die has m sides i.e has values from 1 to m. Given m,n,x calculate the probability that the sum of all the dice is greater than or equal to x.

Example  n=4 m=2 x=8 ans: 1/16=0.0625  n=2 m=6 x=3 ans:35/36=0.972