Digits & Stuff

Let x minus its leading digit be equal to one 29th x:

$x - { \beta 10 }^{ n } = \frac { 1 }{ 29 } x$

$\frac { 28 }{ 29 } x = { \beta 10 }^{ n }$

At this point, it can be seen that the factors of 28 are 4 & 7. Thus, for ${\beta 10}^{n}$ to be divisible by 28, ${\beta}$ must be a multiple of 7. In this case, it would be 7.

$\frac { 28 }{ 29 } x= { 7 \cdot 10 }^{ n }$

$\frac { 4}{ 29 } x ={ 10 }^{ n }$

Since 10 has factors 2 & 5, there must be at least two tens to cancel out the 4 on the left hand side. Thus:

$\frac { 1}{ 29 } x = {25 \cdot 10 }^{ n-2 }$

$x = {725 \cdot 10 }^{ n-2 }$

As mentioned earlier, the smallest amount of tens required to cancel out the 4 was 2, thus by substituting 2 in for n:

$x = 725$

Let $n$ be the number of digits of this number, and $a$ be the first digit, and $b$ the rest of the number. Then, solving for $b$ , we have

$b=\dfrac { 1 }{ 4 } \dfrac { 1 }{ 7 } 10^{ n }a$

Thus, if $b$ is to be an integer and $a$ is a single digit, then $a=7$ , and
$n\ge 2$ , and so we end up with $725$