Straight Scale

NOTE: At the time of my writing (19th Feb 2014), the term 'scale' was used in place of what I presume to be the more accurate term 'base'. I shall be using the latter term in my solution.

Let the number in scale $7$ be $\overline{abc}$ , then the number in base $9$ would conversely be $\overline{cba}$ , where $a,b,c$ are digits (or rather non-negative integers too) not exceeding the value $6$ .

We thus have $49a+7b+c=81c+9b+a$ or equivalently $80c+2b=48a$ , further simplfied to $8(3a-5c)=b$ . Since we have the value of $b$ to be $0 \leq b \leq 6$ , we must have $b=0$ . Consequently, we have $3a=5c$ . Given that $a \neq 0$ (else we wouldn't have a three-digit number in base $7$ ), we must then have $a=5$ and following that $c=3$ .

Our three-digit number in base $7$ is therefore $\fbox{503}$

Each digit in a base- $n$ scale can be represented in the decimal system as a sum of multiple of $n^i$ where $i = 0$ would correspond to the right most digit, $i = 1$ would represent the second digit to the right, and so on. As an explicit example, the septary (base $7$ ) value $111_7$ can be represented in the decimal system as $1 \times 7^2 + 2 \times 7^1 + 3 \times 7^0 = 1 \times 49 + 1 \times 7 + 1 \times 1 = 49 + 14 + 3 = 66$ . Numbers in base- $n$ can only use $n$ digits, so a septary system only uses seven digits from $0$ to $6$ while a nonary (base- $9$ ) system only uses nine digits from $0$ to $8$ .

The problem calls for two equal values in different counting systems, specifically, a septary number $abc_7$ and a nonary number, $cba_9$ , where $a$ , $b$ and $c$ are the same digits. Since both have the same values, their representative values in the decimal system can be equated:

$\begin{aligned} a \times 7^2 + b \times 7^1 + c \times 7^0 &= c \times 9^2 + b \times 9^1 + a \times 9^0 \\ a \times 49 + b \times 7 + c \times 1 &= c \times 81 + b \times 9 + a \times 1 \\ 49a + 7b + c &= 81c + 9b + a \end{aligned}$

As a personal approach to how I solved this (which is definitely longer, there are shorter solutions (refer to Low's)), I isolated all multiples of $a$ on one side of the equation and moved the rest to the other side:

$\begin{aligned} 49a + 7b + c &= 81c + 9b + a \\ 48a &= 2b +80c \\ 24a &= b + 40c \\ a &= \frac{b}{24} + \frac{40c}{24} \\ a &= \frac{b}{24} + \frac{5c}{3} \end{aligned}$

It should be remembered that $a$ , $b$ and $c$ can be only represented by digits from $0$ to $6$ because any digit 'higher' than $6$ cannot be represented in the septary scale, even if it can be in the nonary scale. Also, $a$ , $b$ and $c$ are digits and necessarily they should be represented as whole numbers. For that to happen, $b$ should be a multiple of $24$ and $c$ should be a multiple of $3$ in order that $a$ will surely be a whole number. But, $b$ cannot be $24$ or some higher multiple since we can only admit digits up to $6$ . The only plausible digit for $b$ therefore is $0$ . The equation can be further transformed as:

$\begin{aligned} a &= \frac{0}{24} + \frac{5c}{3} \\ a &= \frac{5c}{3} \end{aligned}$

There are three possible digits for $c$ : $0$ , $3$ and $6$ . If $c = 6$ ,

$\begin{aligned} a &= \frac{5(6)}{3} \\ a &= \frac{30}{3} \\ a &= 10 \end{aligned}$

But, ' $10$ ' cannot be a digit since we can only allow digits up to $6$ . If $c = 0$ ,

$\begin{aligned} a &= \frac{5(0)}{3} \\ a &= \frac{0}{3} \\ a &= 0 \end{aligned}$

which is plausible... except that we will have a string of zeroes, $000_7 = 000_9$ , or equivalently $0_7 = 0_9$ : we only have one-digit values! If $c = 3$ ,

$\begin{aligned} a &= \frac{5(3)}{3} \\ a &= 5 \end{aligned}$

which is plausible. The numbers can be written as $503_7 = 305_9$ . To check if this is true, we calculate:

$\begin{aligned} 49(5) + 7(0) + (3) &= 81(3) + 9(0) + (5) \\ 245 + 0 + 3 &= 243 + 0 + 5 \\ 248 &= 248 \end{aligned}$

The sought answer, in septary, is $\boxed{503_7}$ .