Set of moving last digits

Algebra Level 4

$\begin{array} { l l l l l l l l } & \color{#3D99F6}A & \color{#3D99F6}B & \color{#3D99F6}C & \color{#3D99F6}D & \color{#3D99F6}E & \color{#D61F06}L \\ \times & & & & & & M \\ \hline & \color{#D61F06}L & \color{#3D99F6}A & \color{#3D99F6}B & \color{#3D99F6}C & \color{#3D99F6}D & \color{#3D99F6}E \\ \end{array}$

In the multiplication above, there is a 6-digit positive integer with a last digit $\color{#D61F06}L$ . If its last digit is shifted to the front of the number, while the other 5 digits remain unchanged, then the resulting number is equal to $M$ times the original number - where $M$ is a single-digit positive integer >1.

There are exactly seven different 6-digit integers that exist where this is possible - where $M>1$ - though none of the digits in this problem need to be distinct. (The diagram is just there to help conceptualise the operation of shifting the last digit)

However, one of the seven 6-digit integers is not like the others, and is the odd one out, because it's value of $M$ is different.

If $S$ is the sum of all seven of the original 6-digit numbers, and $O$ is the 6-digit number that is the odd one out.

What is $\large\dfrac{S}{O}$ ?

The answer is 8.

1 solution

Jonathan Quarrie
Jun 2, 2017

Each 6-digit integer can be established using the following equation:

$\color{#D61F06}L \color{#333333}\times 100000 + \color{#3D99F6}\text{x} \color{#333333}= M \times (10\color{#3D99F6}\text{x} \color{#333333}+ \color{#D61F06}L\color{#333333})$

Where $\color{#3D99F6}\text{x}$ represents the 5 unknown, unchanging digits.

For example, where $L = \color{#D61F06}9$ and $M = 4$

$\color{#D61F06}9\color{#333333}00000 + \color{#3D99F6}\text{x} \color{#333333}= 4(10\color{#3D99F6}\text{x}\color{#333333}+\color{#D61F06}9\color{#333333})$
$900000 + \color{#3D99F6}\text{x} \color{#333333}= 40\color{#3D99F6}\text{x}\color{#333333}+36$
$900000 = 39\color{#3D99F6}\text{x}\color{#333333}+36$
$899964 = 39\color{#3D99F6}\text{x}$
$\color{#3D99F6}23076 \color{#333333}= \color{#3D99F6}\text{x}$

So the original 6-digit number is $\color{#3D99F6}23076\color{#D61F06}9$ . And when multiplied by 4, it becomes $\color{#D61F06}9\color{#3D99F6}23076$ .

Repeating the above process for all permutations of single-digit integer values for $\color{#D61F06}L$ and $M>1$ . All valid results produce a 5-digit integer for $\color{#3D99F6}x$ .

Original $\color{#3D99F6}x\color{#D61F06}L$	$\times$	M	=	Result $\color{#D61F06}L\color{#3D99F6}x$
$\color{#3D99F6}23076\color{#D61F06}9$	$\times$	$4$	=	$\color{#D61F06}9\color{#3D99F6}23076$
$\color{#3D99F6}20512\color{#D61F06}8$	$\times$	$4$	=	$\color{#D61F06}8\color{#3D99F6}20512$
$\color{#3D99F6}17948\color{#D61F06}7$	$\times$	$4$	=	$\color{#D61F06}7\color{#3D99F6}17948$
$\color{#3D99F6}15384\color{#D61F06}6$	$\times$	$4$	=	$\color{#D61F06}6\color{#3D99F6}15384$
$\color{#20A900}14285\color{#D61F06}7$	$\times$	$\color{#20A900}5$	=	$\color{#D61F06}7\color{#20A900}14285$
$\color{#3D99F6}12820\color{#D61F06}5$	$\times$	$4$	=	$\color{#D61F06}5\color{#3D99F6}12820$
$\color{#3D99F6}10256\color{#D61F06}4$	$\times$	$4$	=	$\color{#D61F06}4\color{#3D99F6}10256$

$O = \color{#20A900}14285\color{#D61F06}7\color{#333333}$ , because it is the only 6-digit integer from the above set where $M = \color{#20A900}5$ , whereas for all of the others $M = 4$ .

$S = \color{#3D99F6}23076\color{#D61F06}9$ + $\color{#3D99F6}20512\color{#D61F06}8$ + $\color{#3D99F6}17948\color{#D61F06}7$ + $\color{#3D99F6}15384\color{#D61F06}6$ + $\color{#20A900}14285\color{#D61F06}7$ + $\color{#3D99F6}12820\color{#D61F06}5$ + $\color{#3D99F6}10256\color{#D61F06}4 \color{#333333}= 1142856$

$\dfrac{S}{O} = \dfrac{1142856}{\color{#20A900}142857} = \large\boxed{8}$

Set of moving last digits

The answer is 8.

1 solution

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