Loading....
Coupon Accepted Successfully!

 

Addition


Start with the units place digit, 5 + 6 = 11 which is 14
7. So, unit digit is 4 and carry over is 1.
 
Next is tens place digit, 2 + 5 + 1 (carry over) = 8 which is 117. So, tens digit is 1 and carry over is again 1.
 
Next is 3 + 4 + 1 (carry over) = 8 which is 117.

Subtraction

4568  3678

Starting with the units digit, since 6 is smaller than 7, we will borrow 1 from the tens place digit. So, now it is 14 (when the base is 10, we get 10 but here base is 8, so we get 8) and
7 subtracted from it = 14 – 7 = 7, which is the units digit.
 
Next, tens digit is now 4 and we have to subtract 5 from it. We again borrow 1 from hundred’s place digit. So, now it is 12 and 12 – 6 = 6, which is the tens place digit.
 
Now, hundred’s place digit is 3(4  1), so 3  3 = 0
 
Note: Another method of doing these kinds of calculations is to convert these values (in whatever base) into decimal system, then do the actual calculation in decimal system itself and finally converting the numbers into the required or given system.

Some standard systems of writing

  1. Decimal system
     
    Digits used: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
     
    Total digits used = 10 digits
  2. Hexa-decimal system
     
    Digits used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F.
     
    Total digits used = 16
  3. Octal system
     
    Digits used: 0, 1, 2, 3, 4, 5, 6 and 7.
     
    Total digits used = 8
  4. Binary system
     
    Digits used: 0, 1
     
    Total Digits used = 2

Comparison of Different Base Systems

It should be understood here that using different bases creates a difference only in the numbers (more than single digit) and not in the digit. This means in that (5)10 is same as (5)9 or any other system of writing the numbers.
 
Example-1
How many different values of N are possible in the following calculation
(2)N × (4)N = (8)N?
Solution
Any value of N which has 8 as a digit in that system will satisfy this calculation. So, any value of N ≥ 9 is possible. Hence infinite values of N are possible.
 
This concept can be further understood through the mechanism of multiplication. Multiplication is basically the condensed form of addition. So, 2 added up four times will give 8, and hence all the base systems which have the following can be one of the values of N:
0 1 2 3 4 5 6 7 8.
 
 
Example-2
How many different values of N are possible in the following calculation (4)N + (5)N = (9)N?
Solution
In this case also, the digits used can be of any system that has the digit 9. So, all the inegral values N  10 are possible.
 
 
Example-3
How many different values of N are possible in the following calculation. (4)N × (5)N = (24)N?
Solution
Obviously, there can be either just one possible value of N or no possible value of N.
(4)10 × (5)10 = (20)10 [We know (4)N = (4)10 and (5)N = (5)10]
Now, (24)N = 2 × N1 + 4 × N0 = 20
Hence, N = 8. And this will be the only value of N.
 





Test Your Skills Now!
Take a Quiz now
Reviewer Name