# Rearranging Digits – Analytic Solutions

Posted on September 22, 2016

In this post we will work through some digit rearrangement problems.
1. Six-Digit Numbers: The 6-digit number $ABCDEF$ is exactly $23/76$ times the number $BCDEFA$ (where $A$, $B$, $C$, $D$, $E$, and $F$ represent single-digit non-negative integers and $A,B>0$). What is $ABCDEF?$
2. Arithmetic Progression: Find all ordered pairs of distinct single-digit positive integers $(A,B,C)$ such that the three 3-digit positive integers $ABC$, $BCA$, and $CAB$ form an arithmetic progression.
3. Geometric Progression: Find all ordered pairs of distinct single-digit positive integers $(A,B,C)$ such that the three 3-digit positive integers $ABC$, $BCA$, and $CAB$ form a geometric progression.

This post will include analytic solutions, a future post will show some code-based solutions.

#### Six-Digit Numbers

Let's attack the first problem, repeated for clarity:

The 6-digit number $ABCDEF$ is exactly $23/76$ times the number $BCDEFA$ (where $A$, $B$, $C$, $D$, $E$, and $F$ represent single-digit non-negative integers and $A,B>0$). What is $ABCDEF?$

We will let $x=BCDEF$ be a five-digit number. Then

\begin{eqnarray*} ABCDEF &=& \frac{23}{76}\cdot BCDEFA \\ 100000\cdot A+x &=& \frac{23}{76} (10\cdot x+A) \\ 7600000\cdot A+76\cdot x &=& 230\cdot x+23\cdot A \\ 7600000\cdot A+76\cdot x &=& 230\cdot x+23\cdot A \\ 7599977\cdot A &=& 154\cdot x \\ 98701\cdot A &=& 2\cdot x \end{eqnarray*}

Remember, $x$ is a five-digit integer and $A$ is a one-digit number, so one such solution is $A=2$ and $x=98701$. Are there other solutions though? Well, $A$ must be even for $x$ to be an integer, but if $A>2$ then $x$ would be a six-digit number. So there is just the one solution. As a check, this makes our numbers $ABCDEF=298701$ and $BCDEFA=987012$.

(Side note: if you are wondering where these seemingly random six-digit numbers came from, check out the repeating decimals $\frac{23}{77}$ and $\frac{76}{77}$. Fractions whose denominators are factors of $999,999$ will create repeating decimals with cycles of six digits or less and quite often use the same digits in different order.

#### Arithmetic Progression

Now let's look at the second problem, repeated for clarity: Find all ordered pairs of distinct single-digit positive integers $(A,B,C)$ such that the three 3-digit positive integers $ABC$, $BCA$, and $CAB$ form an arithmetic progression.

We will approach this the same way, let's define the 3 numbers \begin{eqnarray*} N_1&=&100A+10B+C \\ N_2&=&100B+10C+A \\ N_3&=&100C+10A+B \end{eqnarray*}

These terms are in arithmetic progression, which can be exploited. Let the common difference of this sequence be $d$, then \begin{eqnarray*} N_2-N_1&=& 100B+10C+A-(100A+10B+C)=d\\ N_3-N_2&=& 100C+10A+B-(100B+10C+A)= d\\ N_3-N_1&=& 100C+10A+B -(100A+10B+C)= 2d \end{eqnarray*}

After simplifying, we have \begin{eqnarray*} -99A +90B +~9C - d=0\\ ~9A -99B +90C - d =0\\ -90A ~-9B +99C - 2d =0 \end{eqnarray*}

We have a system with three equations and four unknowns, which should make you a little anxious (and the third equation is just the sum of the first two). However, if we simplify a little bit (one could do this by hand, but rref in MATLAB is our friend)

Using rref on the coefficient matrix gives us

1  0  -1  0.021021
0  1  -1  0.012012
0  0   0         0

We need integer solutions, so where do we go from here? Take a look at the two decimals, $0.\overline{021}$ and $0.\overline{012}$. The fraction equivalents are $\displaystyle\frac{7}{333}$ and $\displaystyle\frac{4}{333}$ respectively.

The equations are then \begin{eqnarray*} a-c+\frac{7}{333}d=0 \\ b-c+\frac{4}{333}d=0 \end{eqnarray*}

So if we need integers, let's get rid of the fraction by letting $d=333$. Then we have $a-c=-7$, and $b-c=-4$. This gives us the solutions $(a,b,c)=(1,4,8)$ and $(2,5,9)$. Did we get them all though? We can also let $d=-333$, which yields $a-c=7$, and $b-c=4$. This gives us the solutions $(a,b,c)=(8,5,1)$ and $(9,6,2)$.

We could try multiples of $333$ as well, but will quickly run into trouble. If $d=666$, then $a-c=14$, and $b-c=8$ but $a$ and $c$ must be single digit numbers and larger values of $d$ will violate this. We can also let $d=0$ but that just gives us the trivial solutions of an arithmetic progression with common difference $0$.

Therefore our four solutions are \begin{eqnarray*} 148,~481,~814 \\ 185,~518,~851 \\ 259,~592,~925 \\ 296,~629,~962 \\ \end{eqnarray*}

If you are wondering if these numbers came out of nowhere, check out the decimal equivalents of the $27$ths. It is no coincidence that $\displaystyle\frac{4}{27}=0.\overline{148}$, $\displaystyle\frac{13}{27}=0.\overline{481}$, and $\displaystyle\frac{22}{27}=0.\overline{814}$.

#### Geometric Progression

Now let's look at the last problem, again repeated for clarity: Find all ordered pairs of distinct single-digit positive integers $(A,B,C)$ such that the three 3-digit positive integers $ABC$, $BCA$, and $CAB$ form a geometric progression.

For the geomtric sequence \begin{eqnarray*} N_2-r\cdot N_1&=& 100B+10C+A-r(100A+10B+C)= 0\\ N_3-r\cdot N_2&=& 100C+10A+B-r(100B+10C+A)= 0 \end{eqnarray*}

There are other manipulations, but I haven't found any that are linearly independent of the two equations shown.

Not only do we have two equations and four unknowns, the equations are non-linear. With some simplification, we have \begin{eqnarray*} 100B+10C+A-r(100A+10B+C)= 0\\ 100C+10A+B-r(100B+10C+A)= 0 \end{eqnarray*}

Using the symbolic toolbox and rref we can simplify

% a  b                          c
[ 1, 0,     (r - 10)/(10*r^2 - 1)]
[ 0, 1, (r*(r - 10))/(10*r^2 - 1)]


Not incredibly helpful. However, we can reverse the columns of the array (now instead of A,B,C we have C,B,A) and rref gives us

% c  b                      a
[ 1, 0, (10*r^2 - 1)/(r - 10)]
[ 0, 1,                    -r]


Which is a little cleaner. So where do we go from here? Well, from the last equation we have $b-ra=0 \rightarrow r=b/a$ from the second equation. So we could just try every $(a,b)$ combination ($72$ since $a\neq b$) and see if $$c=\displaystyle\frac{10r^2 - 1}{10 - r}a$$ is an integer. Still some brute force required though. You could also substitute $r=b/a$ into $c=\displaystyle\frac{10r^2 - 1}{10 - r}a$ yielding $$c= \displaystyle\frac{10b^2-a^2}{10a - b}$$ which we just have to ensure is an integer. But again, it looks like guessing and checking is required. You could even simplify further with some polynomial division $$c = \displaystyle\frac{10b^2-a^2}{10a - b} = \displaystyle\frac{999a^2}{10a - b} - 10b - 100a$$ which takes you down a path of finding $a$ and $b$ such that $10a - b$ is a divisor of $999a^2$, but that may even be more dangerous than guess and checking since it's so easy to leave out a case and every case must be checked to ensure $c$ is a positive single digit integer.

You can though, and it will lead you to two ordered triples, $(a,b,c)=(4,3,2)$ and $(8,6,4)$ (the second is just a less exciting multiple of the first, both have the common ratio $r=\frac{3}{4}$. The geometric series are $432,~324,~243$ and $864,~648,~486$.

This last solution left me feeling less than satisfied. In the next post we will do a code-based solution to ensure with some certainty that those are the only values.

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