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By R.S. Sedha

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625 Ans. Example 2-39. Covert the following into Binary and Hexadecimal (a) 243 (b) 534 (Mahatma Gandhi; University, May 2003) Solution. (a) Given the decimal number = 243 Decimal-to-Binary We know that the conversion from decimal-to-binary can be done using repeated division by 2 method. Top 243 ÷ 2 121 ÷ 2 60 ÷ 2 30 ÷ 2 15 ÷ 2 7÷2 3÷2 1÷2 = = = = = = = = 121 60 30 15 7 3 1 0 with a remainder with a remainder with a remainder with a remainder with a remainder with a remainder with a remainder with a remainder 1 1 0 0 1 1 1 1 (LSB) (MSB) Bottom Reading the remainder from bottom to top the binary number for the decimal number 243 is 11110011  24310 = 111100112 Ans.

4448 with a carry of 0 LSB MSB Bottom and so on……… We can continue this process further if we wish because the binary fractional part has not reduced to zero. However we stop here because the desired number of binary places is reached. 1110. 11102 Ans. Example 2-14. Obtain the binary equivalent of 15 and 25. (Gauhati University, 2007) Solution. Binary equivalent of 15 Given: The decimal number = 15 We know that the decimal number can be expressed by the sum-of-weight as follows. 15 = 8 + 4 + 2 + 1 = 23 + 22 + 21 + 20 Placing 1s in the appropriate positions, 23, 22, 21 and 20 position, Thus 23 22 21 20 1 1 1 1  1510 = 11112 Ans.

Step 2. Divide each resulting quotient by 16 until there is a zero whole number quotient. Step 3. The remainders generated by each division form the hexadecimal number. The first remainder to be produced is the least-significant digit (LSD) in the hexadecimal number and the last remainder to be produced is the most-significant digit (MSD). In other words, reading the remainders from bottom-to-top constitutes the hexadecimal number. Note that any remainders that are greater than 9 are represented by the letters A through F.

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