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)

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).

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

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 ?

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

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