This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.

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Okay, let's think about this like we're using blocks or counting on our fingers, just like primary pupils learn.
First, we need to understand what binary numbers mean in our normal counting system (base 10). In binary, each place value is double the one before it, starting from on the right.
Step 1: Change the first binary number () into our normal number.
We have: • One group of () • One group of () • One group of () • Zero groups of ()
Add these up: . So, is the same as .
Step 2: Change the second binary number () into our normal number.
We have: • One group of () • One group of () • One group of ()
Add these up: . So, is the same as .
Step 3: Add the two normal numbers together. Now we just add . .
Step 4: Change our answer () back into a binary number. We need to find which binary place values add up to . We'll use the place values: .
• Does fit into ? Yes. So we use one . left over. Binary: (for the s place)
• Does fit into ? No. So we don't use an . Binary: (for the s place)
• Does fit into ? Yes. So we use one . left over. Binary: (for the s place)
• Does fit into ? No. So we don't use a . Binary: (for the s place)
• Does fit into ? Yes. So we use one . left over. Binary: (for the s place)
Putting the binary digits together from left to right (from the s place down to the s place), we get .
The final answer is .
3 done, 2 left today. You're making progress.
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Okay, let's think about this like we're using blocks or counting on our fingers, just like primary pupils learn.
This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.