Base Number Arithmetic

Arithmetic in different bases works exactly the same, but takes some getting used to. It’s easy to forget when to carry and make compounding mistakes.

Addition

Addition in other bases works the exact same as base 10. We could either add the numbers directly,

21_6 + 14_6 = 25_6 + 10_6 = 35_6

Or convert to base 10 and preform our addition normally

21_6 + 14_6 = 13_{10} + 10_{10} = 23_{10} = 35_6

Subtraction

Subtraction unsuprisingly is no different. In the example below we seperate the number into groups of 9.

\begin{align*} 517_9 - 214_9 &= (5 - 2) \cdot 9^2 + (1 - 1) \cdot 9^1 + (7 - 4) \cdot 9^0\\ &= 3 \cdot 9^2 + 0 \cdot 9^1 + 3 \cdot 9^0\\ &= 303_9 \end{align*}

Multiplication

When multiplying two numbers of any base, it is almost always easier to regroup the multiplicands into bundles.

\begin{align*} 435_6 &= 4 \cdot 6^2 + 3 \cdot 6^1 + 5 \cdot 6^0\\ 12_6 &= 0 \cdot 6^2 + 1 \cdot 6^1 + 2 \cdot 6^0 \end{align*}

We can then multiply the numbers using normal methods

(4 \cdot 36 + 3 \cdot 6 + 5) \cdot (6 + 2) = \\ = 167 \cdot 8\\ = 1336

Division

The process of long division is the same in every base, so I won’t waste your time with needless examples (I don’t wanna Tex them).

Problems

Review

Alcumus - Level 18 (#29673)

Find the positive base b in which the equation 13\cdot15=243 is valid.

By simply multiplying the equation out, we get a quadratic

(b + 3)(b + 5) = 2b^2 + 4b + 3

Simplifying to

(b - 6)(b + 2) = 0

Since b must be positive, b must be equal to 6.

Challenge

Alcumus - Level 23 (#29671)

If S, H, and E are all distinct non-zero digits less than 5 and the following is true, find the sum of the three values S, H, and E, expressing your answer in base 5.

\begin{array}{cccc} &S&H&E_5\\ &+&H&E_5\\ \hline &S&E&S_5\\ \end{array}

Removing the leftmost digit gives us

\begin{array}{cccc} &&H&E_5\\ &+&H&E_5\\ \hline &&E&S_5\\ \end{array}

Assuming there is no carry on the rightmost digit, we have S = 2E and E = 2H. After some algebra we arrive at S = 4, H = 1, E = 2. Thus we have the sum 12_5