\((110010000)_2\)
To find the sum of two binary numbers, we perform binary addition similarly to decimal addition, with the only difference being that in binary, the digits are 0 and 1.
| Binary Addition Rules: | |
|---|---|
| \(0 + 0 = 0\) | No carry |
| \(0 + 1 = 1\) | No carry |
| \(1 + 0 = 1\) | No carry |
| \(1 + 1 = 0\) | Carry 1 |
To add \((101101101)_2\) and \((100011)_2\):
- Align the numbers from the right:
- 101101101 + 000100011 ---------
- Start adding from the rightmost bit:
- Below is the step-by-step addition:
1 (carry from next step) 101101101 + 000100011 ------------ 110010000
Explanation of each step:
- The rightmost digits are 1 and 1. According to our rules: \(1 + 1 = 0\) with a carry of 1.
- Next, add 1 (from second last bit of the upper number) and 1 (carry), resulting in \(0\), with another carry over.
- Continue the addition to the left; when no further carry is required, just add the digits normally.
- Finally, add the remaining digits, taking care of each carry step.
Hence, the sum of \((101101101)_2\) and \((100011)_2\) is \((110010000)_2\).
Correct Answer: \((110010000)_2\).
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Important Questions from Number System
What is the value of 1 2 + 2 2 + 3 2 + ......21 2 ?
Which sequence is correct to represent the hierarchical chain of number system?
(Where N - Natural Numbers
W - Whole Numbers
Q - Rational Numbers
Z - Integers)
What must be added to 45680 to make it exactly divisible by 9?
How many zeroes are there at the end of the following product?
1 x 5 x 10 x 15 x 20 x 25 x 30 x 35 x 40 x 45 x 50 x 55 x 60
Let XYZ be a three-digit number, where (x + y + Z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by
