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A special number system (sns ) with a range -3 to +3 having the values { C,B,A,0,1,2,3 } Tell the base of the question. I know how to convert from 100 base 10 to _ _ base 8 which 144. But I want to convert. 100 base 10 to _ _ base sns. I thought base is of 4 and I got answer as 1110 but I want it in the sns format Please help me in solving

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Your "special" number system is known as a balanced base-seven number system. It is "base seven" because there are seven values a digit can have ($-3$ through $3,$ inclusive), and it is "balanced" because for each positive digit value there is a negative digit value of equal magnitude.

I prefer to write the names of the digits as $\{\bar1,\bar2,\bar3,0,1,2,3\}$ because the correspondence between positive and negative values is clearer. (Also, this notation wouldn't get in the way if I wanted to work with balanced base-fifteen.)

I will assume you are familiar with the method that uses by division by $7$ with remainder to convert a base-ten number to "ordinary" base seven, but for review, let's convert $\newcommand{ten}{\mathrm{ten}}80_\ten$ to base seven by this method.

Step 1: Divide $80_\ten$ by $7,$ producing a quotient of $11_\ten$ and a remainder of $3.$ So the rightmost digit of the result is $3.$

Step 2: Take the quotient from the previous step, $11_\ten,$ and divide by $7.$ The new quotient is $1$ and the remainder is $4,$ so the we put a $4$ on the left side of the result, which is now $43.$

Step 3: We have $1$ from the previous step, divided by $7$ gives a quotient of $0$ and a remainder of $1,$ so we write $1$ on the left side of the result, which is now $143.$ Since the quotient in this step was zero, $\newcommand{seven}{\mathrm{seven}}143_\seven$ is the final answer.

A more compact notation for this division-with-remainder example is: \begin{align} 80 &= 7 \times 11 + 3, \\ 11 &= 7 \times 1 + 4, \\ 1 &= 7\times 0 + 1 \quad \longrightarrow 143_\seven. \end{align}

The reason this works is that the last digit of a number $N$ written in base $b$ is always congruent to $N$ modulo $b.$ Division by $b$ with remainder gives us that number.

For balanced base seven, we also divide by seven, but with a twist: the "remainder" has to be between $-3$ and $3,$ inclusive. That means sometimes we will take "one too many $7$s" from a number, leaving a negative remainder.

To write $80_\ten$ in balanced base seven:

Step 1. Divide $80_\ten$ by $7.$ As before, the quotient is $11_\ten$ and the remainder is $3,$ which is the rightmost digit of the result.

Step 2. Divide $11_\ten$ by $7.$ The usual way gives a remainder of $4,$ which is too large, so we increase the quotient by $1$: the result is a quotient of $2,$ and remainder $11_\ten - 2\times7 = -3,$ which gives us the digit $\bar3.$

Step 3. Divide $2$ by $7,$ for a quotient $0$ and remainder $2.$

In compact notation: \begin{align} 80 &= 7 \times 11 + 3, \\ 11 &= 7 \times 2 - 3, \\ 2 &= 7\times 0 + 2 \quad \longrightarrow 2\bar33_\mathrm{seven,bal}. \end{align}

Another example: convert $1500_\ten$ to balanced base seven.

\begin{align} 1500 &= 7 \times 214 + 2, \\ 214 &= 7 \times 31 - 3, \\ 31 &= 7 \times 4 + 3, \\ 4 &= 7\times 1 - 3, \\ 1 &= 7\times 0 + 1 \quad \longrightarrow 1\bar33\bar32_\mathrm{seven,bal}. \end{align}


If you already have a number in base seven, conversion to balanced base seven is relatively easy. Starting on the right and working to the left, if any digit is greater than $3$ you subtract $7$ and carry $1$ into the next digit to the left.

For example, to convert $143_\seven,$ the $3$ can stay as it is; $4$ is too large, so you subtract $7$ to get $\bar3$ and carry $1$ into the next digit, which therefore becomes $1+1=2.$ Since $2$ is not too large, you leave it alone, and you're done.

A more formal description of this procedure is given in the answer linked here.

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  • $\begingroup$ i was unable to understand the answer given. in the link $\endgroup$ – pullarao kommineni Feb 10 '17 at 5:13
  • $\begingroup$ I have added notes on how to convert directly from base 10 to balanced base 7, which may be preferable to using base 7 as an intermediate step. $\endgroup$ – David K Feb 11 '17 at 2:44
  • $\begingroup$ thank u sir for the great explanation $\endgroup$ – pullarao kommineni Feb 11 '17 at 6:14

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