JOMO 5, Short 1

$m \equiv 5 \enspace (\mod m-5 )$ $$

$m^5 \equiv 5^5 \enspace (\mod m-5 )$ $$

$m^5 + 5^5 \equiv 2 \times 5^5 \enspace (\mod m-5 )$

$$

We know, $m^5 + 5^5 \equiv 0 \enspace (\mod m-5 )$ $$

$(m-5)$ divides $2\times5^5$ . The greatest value of $(m-5)$ can be $2\times5^5$ .

$$

Hence $m$ 's greatest value = $2\times5^5+5=6255$ . Last $3$ digits $\boxed{255}$

If we carry out polynomial division on $\frac{x^5+5^5}{x-5}$ we get the remainder $2\times5^5$ . Therefore by the division theorem the remainder has to be divisible by $m-5$ for there to be no remainder. The largest value of $m-5$ that divides $2\times5^5$ would be $m-5=2\times5^5$ and by solving for $m$ we get $m=2\times5^5+5=6255$ thus the last 3 digits are $\boxed{255}$ .

Very similar to @Ali Caglayan . But I'm posting it anyway.

$\dfrac{m^5+5^5}{m-5}$

$=\dfrac{m^5-5^5+2\times 5^5}{m-5}$

$=\dfrac{m^5-5^5}{m-5}+\dfrac{2\times 5^5}{m-5}$

$=k+\dfrac{2\times 5^5}{m-5}\quad [\text{since}(a-b)\mid (a^n-b^n)]$

If we want to maximize $m$ , we have to minimize that fraction [but still make sure that it is equal to an integer]. The minimum (positive) integer value of that fraction is $1$ and for that to happen $m$ has to equal $2\times5^5+5=6255$ .

So its last three digits are $\boxed{255}$ .