Question

add in the indicated base.

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$$\begin{array}{r} 212_{\\text{five}} \\\\ +\\ 443_{\\text{five}} \\\\ \\hline \\end{array}$$

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the sum is \\(\square_{\text{five}}\\).

Explanation:

Identify the given values and operation

We are given two numbers in base five to add:

$$ 212_{\text{five}} + 443_{\text{five}} $$

Add the units column (rightmost digit)

We add the digits in the units place:

$$ 2 + 3 = 5 $$

Using Place Value Exchanges, since we are in base five, a group of \(5\) is carried over as \(1\) to the next column:

$$ 5 = 10_{\text{five}} $$

The units digit is \(0\), and we carry over \(1\).

Add the fives column (middle digit)

We add the digits in the fives place along with the carryover:

$$ 1 (\text{carry}) + 1 + 4 = 6 $$

Using Place Value Exchanges, we convert \(6\) to base five:

$$ 6 = 11_{\text{five}} $$

The fives digit is \(1\), and we carry over \(1\).

Add the twenty-fives column (leftmost digit)

We add the digits in the twenty-fives place along with the carryover:

$$ 1 (\text{carry}) + 2 + 4 = 7 $$

Using Place Value Exchanges, we convert \(7\) to base five:

$$ 7 = 12_{\text{five}} $$

The twenty-fives digit is \(2\), and we carry over \(1\) to the next place.

Combine the results

Combining all the digits from left to right gives:

$$ 1210_{\text{five}} $$

Answer:

Add in the indicated base.

$$
LATEXBLOCK0
$$

The sum is <blank>\(1210\)</blank>\(_{\text{five}}\).