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I'm trying to understand number systems. Consider a base-52 number system consisted of following symbols as digits:

abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ

synonymous to 0123456789 digits in the base-10 number system.

how would we convert 100 from base-10 to base-52 ?

dsfx3d
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  • $100=52+48$. Therefore, it is $bW$. – GReyes Mar 11 '19 at 07:06
  • @GReyes thanks, can you explain how you add 52 and 48 in base-52 – dsfx3d Mar 11 '19 at 07:20
  • it should be aW then – dEmigOd Mar 11 '19 at 07:23
  • http://www.oxfordmathcenter.com/drupal7/node/18 – Matti P. Mar 11 '19 at 07:27
  • Essentially you just take the modulus of the number (in base 52) repeatedly. That's of course a simplification, but it's essentially what you do. – Matti P. Mar 11 '19 at 07:28
  • I understand that $a$ corresponds to $0$. The thing is that $100=52+48$ (as numbers, irrespective of their notation in base $10$). The digit $a$ means "zero times" the base to the corresponding power, $b$ means "once the base to the corresponding power" etc. $100$ is "once the base" plus 48 units. These 48 units are represented by $W$ in base $52$. Another example: 7 in base $2$ is 111, because it is "once" $2^2$ plus "once" $2$ plus one unit. In this case the digits are represented using the usual symbols, but you could use, say, $a$ and $b$ instead. Then $7$ would be $bb$. – GReyes Mar 11 '19 at 07:34

3 Answers3

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Just as the decimal number $100$ denotes $1 \cdot 10^2 + 0 \cdot 10^1 + 0 \cdot 10^0$, we can express $100$ in base $52$ by writing it as a linear combination of powers of $52$ (all of whose coefficients are between $0$ and $52 - 1 = 51$). Dividing by $52$, we see that $100 = 1 \cdot 52^1 + 48 \cdot 52^0$, so the base $52$-representation has the $48$th symbol (counting $a$ as the $0$th), namely $\texttt{W}$, in the rightmost place, and the $1$st symbol, namely $\texttt{b}$, in the second-rightmost place. Thus, using subscripts to denote bases for emphasis, we have $$\boxed{100_{10} = {\texttt{bW}}_{52}} .$$

Travis Willse
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An answer using example: To convert 141548 to base 52.

Divide by 52 and write remainder in terms of the given digits:

$$\begin{array}{rlc} 141548\div 52 &= 2722\ rem\ 4& 4\to e\\ 2722 \div 52 &= 52\ rem\ 18&18\to s\\ 52\div 52&=1\ rem\ 0&0\to a\\ 1\div 52&=0\ rem\ 1&1\to b \end{array}$$ Hence the number is written as "base".

This says that $141548=b\times 52^3+a\times 52^2+s\times 52+e=((b52+a)52+s)52+e$.

Chrystomath
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Here's the third recommendation of counting in base-52:

0, a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z, A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, a0, aa, ab, ac, ad, ae, af, ag, ah, ai, aj, ak, al, am, an, ao, ap, aq, ar, as, at, au, av, aw, ax, ay, az, aA, aB, aC, aD, aE, aF, aG, aH, aI, aJ, aK, aL, aM, aN, aO, aP, aQ, aR, aS, aT, aU, aV, aW, aX, aY, b0, ba, ...

(The zero's do need a font that puts a diagonal line through them. The zero's flag the group counts and the zero's allow for a sensible addition operation.)

Then

(52 + 48) base-10 = (a0 + V) base-52 = aV base-52

And for instance

(54 + 25) base-10 = (52 + 27) base-10 = (a0 + A) base-52 = aA base-52 .

S Spring
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