A number theory problem by Bryan Lee Shi Yang

The maximum value of $\overline{abcdef}$ in order to retain only $6$ digits when multiplied by $6$ is $166666$ , leading to $a=1$ .

It's not too hard to prove that the first number must always increase when multiplied by $2,3,4,5$ and $6$ , so the first number of each multiplication forms a sequence of six increasing numbers, starting with $1$ , meaning that all the digits of $\overline{abcdef}$ must be different (and greater than or equal to $1$ ). This rules out $0$ from being any of the digits.

Multiplying an even number by $5$ results in an ending $0$ , and thus $f$ must be odd. We can also rule out $f=5$ for the same reason.

Now let's consider the last digits of $f=1$ , $f=3$ and $f=9$ :

For $f=1$ , the final digits following the respective multiplications are $1,2,3,4,5$ and $6$ . But $a=1$ and $f=1$ , leaving $2,3,4,5$ and $6$ for $4$ remaining digits, which is an impossibility.

Applying the same logic yields last digits of $3,6,9,2,5$ and $8$ for $f=1$ and $9,8,7,6,5$ and $4$ for $f=9$ , both of which produce too many distinct digits.

Therefore $f=7$ and the digits of $\overline{abcdef}$ must be $1, 2, 4, 5, 7$ and $8$ in some order. We can see that the first digit of the original is $1$ , the first digit after multiplication by $2$ is $2$ , the first digit after multiplication by $3$ is $4$ , the first digit after multiplication by $4$ is $5$ etc etc

This tells us that $b$ cannot exceed $4$ (seeing as that would cause the first digit after multiplication by $2$ to be a $3$ ), so $b=2$ or $b=4$ . But $b$ cannot be $2$ because then the maximum value of $\overline{abcdef}$ is $128547$ ; when this is multiplied by $5$ we cannot obtain the first digit of $7$ , we fall short at six-hundred thousand. So $b=4$ .

Thus, the remaining solutions are $148527, 148257, 145827, 145287, 142857$ and $142587$ . Of these, only $142857$ is valid, and we can check this by computing the $5$ operations:

$142857\times 2=285714$

$142857\times 3=428571$

$142857\times 4=571428$

$142857\times 5=714285$

$142857\times 6=857142$

Interestingly, this is the recurrence pattern of $\frac{1}{7}=0.142857142857...$ , and probably the most widely known cyclic number .

Answer is a cyclic number connected to 7.
https://en.wikipedia.org/wiki/Cyclic_number