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Compute 221_3 + 122_3 in base 3.

 Jun 13, 2024
 #1
avatar+1075 
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I'm not great at direct addition, so let's convert everything to base 10. We have

\((221)_3 = (2 × 3^2) + (2 × 3^1) + (1 × 3^0) = (25)_{10} \)

\((122)_3 = (1 × 3^2) + (2 × 3^1) + (2 × 3^0) = (17)_{10} \)

 

Adding these two numbers up, we get \(25+17=42\)

 

So now we want to convert 42 back to base 3. 

First, we do 42/3, which has a quotient of 14 and remainder of 0.

Next, we take 14 and have 14/3, which has a quotient of 4 and remainder of 2. 

Next, we do 4/3, which has a quotient of 1 and remainder 1. 

Lastly, we do 1/3, which has quotient 0 and remainder 1. 

 

So our number is \(42_{10}=1120_3\)

 

Our final answer is \(1120_3\)

 

Thanks! :)

 Jun 13, 2024

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