Number System – 9

Number System – Exercise – 1 (Answer)

1. Divisor = 41, dividend = 15625, quotient = 381 and remainder = 4.

2. Divisor = (dividend – remainder )/quotient = (397246 – 211)/865 = 459.

3. Divisor = (dividend – remainder )/quotient = (x + ak) – x/a = k.

4. On dividing 87375 by 698, the remainder is 125, the least number to be subtracted from the dividend in the remainder

So, the least number to be subtracted = 125.

5. On dividing 49123 by 263, the remainder is 205, the least number to be added to the dividend = divisor – remainder = 263 – 205 = 58.

So, the least number to be added = 58.

6. The greatest 3 digit number = 999, on dividing 999 by 35, remainder = 19.

So, the required number = dividend – remainder = 999 – 19 = 980.

7. The least number of 3-digits = 100. On dividing 100 by 14, remainder = 2

So, the required number = dividend + (divisor – remainder) = 100 + (14 – 2) = 112.

8. Here, the divisor 602 is multiple of the divisor 14, and greater divisor remainder is greatest then smaller divisor.

So, required number = 34/14 = 8.

9. Here, the divisor 357 is the multiple of the divisor 17, but the greater divisor remainder is smaller then smaller divisor,

So, the remainder is same which is 5.

10. The binary equivalent of these number is:-

(i) (30)¹⁰ = (11110)²

(ii) (27)¹⁰ = (11011)²

(iii) (41)¹⁰ = (101001)²

Number System 8

Number System – Exercise – 1

1. On dividing 15624 by 41, what is the quotient and the remainder?

2. On dividing 397246 by a certain number, the quotient is 865 and the remainder is 211. Find the divisor.

3. What is the number which on dividing (x + ak) gives ‘a’ as the quotient and x as the remainder?

4. Find the least number, that must be subtracted from 87375, to get a number exactly divisible by 698.

5. What least number must added to 49123 to get a number exactly divisible by 263.

6. Find the greatest number of 3 digits, which is exactly divisible by 35.

7. Find the least number of 3 digits, which is exactly divisible by 14.

8. A number when divided by 602 leaves a remainder 36. What remainder would be obtained by dividing the same number by 14?

9. A number, when divided by 357, leaves a remainder 5. What remainder would be obtained by dividing the same number by 17?

10. Find the binary equivalent of

(i) 30

(ii) 27

(iii) 41

Number System 7

Binary Number System

This number system has base 2 and use only 0 and 1; whereas the conventional decimal system having a base 10, uses 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

The positional weight of a binary number:-

Positions from right to left	8th 7		7th 6		6th 5	5th 4	4th 3	3rd 2	2nd 1	1st 0
Positional weight of a Binary number	27 128	26 64		25 32		24 16	23 8	22 4	21 2	20 1

By using this table:-

The decimal equivalents of any binary number can be found out.

(10101)₂ = 1 x 2⁴ + 0 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2⁰

= 16 + 0 + 4 + 0 + 1 = (21)₁₀

Alternatively,

1	0	1	0	1
16	8	4	2	1

Hence, (10101)₂ = 16 + 4 + 1= (21)₁₀

Conversion of a Decimal Number to Binary Number

To convert a decimal number to a binary number, the following steps are to be considered.

Step 1:- Divide the decimal number by 2.

Step 2:- Go on dividing the quotients (obtained at each stage) by 2 till the quotient is 0.

Step 3:- Write down the remainders on the right side after each of the above divisions.

Step 4:- Arrange the remainders ( as obtained in Step 3) in the reverse order to get the equivalent binary number.

Ex.:- convert a decimal number (25)₁₀ to the equivalent Binary number.

Soln.:- Divide the given number 25 by 2 and go on dividing the quotient by 2 till the quotient is 0.

Then we get the remainder in every step like 1, 0, 0, 1, 1 respectively.

Now, arrange the remainder in reverse order like, 11001. This is the required Binary number.

So, (25)₁₀ = (11001)₂

Calculation in the Binary System

Mathematical calculations (i.e. addition, subtraction and multiplication) in the binary system follow their own rules and are similar to those in the decimal system.

Binary Addition

It is easy to add two binary numbers. The rules for binary addition are as follows:-

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 10 (put 0 to the same column and carry 1 to the next left column)

Binary Subtraction

It is easy to subtract a binary number from another binary number. The rules for binary subtraction are as follows:

0 – 0 = 0

1 – 1 = 0

1 – 0 = 1

To find 0 – 1, we write 1 in the result and also we borrow 1 from the next left column.

Binary Multiplication

Binary multiplication is similar as decimal multiplication. The four rules that are followed in multiplication of two binary numbers are following:-

0 x 0 = 0

0 x 1 = 0

1 x 0 = 0

1 x 1 = 1

Binary Division

In the division, the method that is applied is similar to that in decimal system. The two rules which are followed here are,

⁰/₁ = 0 and ¹/₁ = 1

Here also, the value of ¹/₀ is undefined.