Problem

Write 41 in the base-four system. \[ 41=\square_{\text {four }} \]

Solution

Step 1 :Find the highest power of 4 that is less than or equal to 41, which is \(4^2=16\).

Step 2 :Divide 41 by 16 to get 2 with a remainder of 9. So, the coefficient for the \(4^2\) term is 2.

Step 3 :Look at the remainder, 9. The highest power of 4 that is less than or equal to 9 is \(4^1=4\).

Step 4 :Divide 9 by 4 to get 2 with a remainder of 1. So, the coefficient for the \(4^1\) term is 2.

Step 5 :The remainder 1 is less than 4, so it is the coefficient for the \(4^0\) term.

Step 6 :Therefore, 41 in base-four is \(\boxed{221_{\text {four }}}\).

Step 7 :To check our work, we can convert \(221_{\text {four }}\) back to base-10: \(2\cdot4^2 + 2\cdot4^1 + 1\cdot4^0 = 32 + 8 + 1 = 41\), which matches the original number.

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