Let m = (313) 4and n = (322) 4. Find the base 4 expansion of m + n.

Question

Prepp Team · Accepted Answer

Understanding Base 4 Numbers

In number systems with a base (like base 4), the position of a digit determines its value, which is a power of the base. Base 4 uses only the digits 0, 1, 2, and 3. When adding numbers in base 4, if the sum of digits in a column is 4 or more, we regroup (carry over) groups of 4 to the next higher place value, just like we regroup groups of 10 in base 10 (decimal) addition.

Adding Numbers in Base 4

The question asks us to find the sum of two base 4 numbers, m and n, where $m = (313)_4$ and $n = (322)_4$. We need to perform addition in base 4. Let's add these numbers column by column, starting from the rightmost digit.

We can set up the addition vertically:

$$ \begin{array}{@{}c@{\,}c@{}c@{}c} & 3 & 1 & 3_4 \\ + & 3 & 2 & 2_4 \\ \hline \end{array} $$

Now, let's add each column, moving from right to left:

Rightmost Column (40 place): Add the digits in the rightmost column: $3 + 2 = 5$. In base 4, the number 5 is equivalent to one group of 4 and a remainder of 1. So, $5 = 1 \times 4 + 1$. We write down 1 in this column and carry over 1 to the next column to the left.
Middle Column (41 place): Add the digits in the middle column plus the carry-over from the previous column: $1 (\text{carry}) + 1 + 2 = 4$. In base 4, the number 4 is equivalent to one group of 4 and a remainder of 0. So, $4 = 1 \times 4 + 0$. We write down 0 in this column and carry over 1 to the next column to the left.
Leftmost Column (42 place): Add the digits in the leftmost column plus the carry-over from the previous column: $1 (\text{carry}) + 3 + 3 = 7$. In base 4, the number 7 is equivalent to one group of 4 and a remainder of 3. So, $7 = 1 \times 4 + 3$. We write down 3 in this column and carry over 1 to the next column to the left.
Next Left Column (43 place): Since there are no more digits in the original numbers, the carry-over 1 becomes the digit in this new leftmost column. We write down 1.

Let's show the addition with carries:

$$ \begin{array}{@{}c@{\,}c@{}c@{}c@{}c} & \stackrel{1}{3} & \stackrel{1}{1} & 3_4 \\ + & & 3 & 2 & 2_4 \\ \hline 1 & 3 & 0 & 1_4 \\ \end{array} $$

So, the sum of $(313)_4$ and $(322)_4$ in base 4 is $(1301)_4$.

Verifying the Base 4 Addition

We can convert the base 4 numbers to base 10 (decimal) to verify the result.

$m = (313)_4 = 3 \times 4^2 + 1 \times 4^1 + 3 \times 4^0 = 3 \times 16 + 1 \times 4 + 3 \times 1 = 48 + 4 + 3 = 55_{10}$

$n = (322)_4 = 3 \times 4^2 + 2 \times 4^1 + 2 \times 4^0 = 3 \times 16 + 2 \times 4 + 2 \times 1 = 48 + 8 + 2 = 58_{10}$

In base 10, the sum is $55 + 58 = 113_{10}$.

Now let's convert our base 4 result $(1301)_4$ to base 10:

$(1301)_4 = 1 \times 4^3 + 3 \times 4^2 + 0 \times 4^1 + 1 \times 4^0 = 1 \times 64 + 3 \times 16 + 0 \times 4 + 1 \times 1 = 64 + 48 + 0 + 1 = 113_{10}$

Since $113_{10} = 113_{10}$, our base 4 addition is correct.

Summary of Base 4 Addition Process

Place Value (Base 4)	Digits to Add	Sum (Base 10)	Sum (Base 4)	Digit in Result	Carry-over
4<sup>0</sup>	3 + 2	5	$(11)_4$	1	1
4<sup>1</sup>	1 (carry) + 1 + 2	4	$(10)_4$	0	1
4<sup>2</sup>	1 (carry) + 3 + 3	7	$(13)_4$	3	1
4<sup>3</sup>	1 (carry)	1	$(1)_4$	1	0

The resulting digits from bottom to top are 1, 3, 0, 1, giving us $(1301)_4$.

Revision Table: Number Bases

Base	Digits Used	Example (Base 10 Conversion)
Base 10 (Decimal)	0, 1, 2, 3, 4, 5, 6, 7, 8, 9	$123_{10} = 1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0$
Base 2 (Binary)	0, 1	$1101_2 = 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 8 + 4 + 0 + 1 = 13_{10}$
Base 4	0, 1, 2, 3	$313_4 = 3 \times 4^2 + 1 \times 4^1 + 3 \times 4^0 = 48 + 4 + 3 = 55_{10}$
Base 8 (Octal)	0, 1, 2, 3, 4, 5, 6, 7	$175_8 = 1 \times 8^2 + 7 \times 8^1 + 5 \times 8^0 = 64 + 56 + 5 = 125_{10}$
Base 16 (Hexadecimal)	0-9, A-F	$2A_{16} = 2 \times 16^1 + 10 \times 16^0 = 32 + 10 = 42_{10}$

Additional Information on Base 4 Arithmetic

Base 4, also known as quaternary, is a number system with four as its base. It is sometimes used in computing contexts, although binary (base 2) and hexadecimal (base 16) are more common. Arithmetic operations like addition, subtraction, multiplication, and division can be performed in base 4 using similar principles to decimal arithmetic, but with regrouping based on powers of 4.

Key points about Base 4:

Only digits 0, 1, 2, and 3 are used.
Each position value increases by a power of 4 as you move left.
When adding, a sum of 4 means a carry-over of 1 to the next place value (since $4 = 1 \times 4^1 + 0 \times 4^0 = (10)_4$).
When subtracting, you "borrow" a 4 from the next higher place value (since $1 \times 4^n = 4 \times 4^{n-1}$).