Answer to Question #110843 in Electric Circuits for DD

Question

Divide 0001010001001001 (BCD) by 10012 and express the result in octal equivalent.

Answer 1

Split the code into 4-bit parts:

$$0001\space0100\space 0100\space 1001\space (BCD) = 1449_{10}.$$ $$1001_2 =1\cdot2^3+0\cdot2^2+0\cdot2^1+1\cdot2^0=9_{10}.$$

Divide:

$$1449_{10}/9_{10}=161_{10}.$$

Convert to octal equivalent.

The remainder will be the last smallest number:

$$\bigg\{\frac{161}{8}\bigg\}=1.$$

The integer part is 20, divide by 8:

$$\bigg\{\frac{20}{8}\bigg\}=4.$$

The integer part is 2, divide by 8 (does not have integer part). We have 2:

$$\bigg\{\frac{2}{8}\bigg\}=2.$$

Now join all our numbers from top to bottom, hence:

$$161_{10}=241_8.$$