Problem source

Find the sum of those numbers between 1000 and 6000 every one of whose digits
is one of the numbers $0,\,2,\,5$ or $7$, giving
 your answer as a product of primes.

Solution source

The first digit is $2$ or $5$, all the other digits can be any value from $0,2,5,7$.

Therefore we have

\begin{align*}
S &= 2000 \cdot 4^3+5000 \cdot 4^3 + (200+500+700) \cdot 2 \cdot 4^2 + (20+50+70) \cdot 2 \cdot 4^2 + (2+5+7) \cdot 2 \cdot 4^2 \\
&= 7 \cdot 4^3 \cdot 2^3 \cdot 5^3 + 14 \cdot 2 \cdot 4^2 \cdot 111 \\
&= 2^{9} \cdot5^3 \cdot 7 + 2^{6} \cdot 3 \cdot 7 \cdot 37 \\
&= 2^6 \cdot 7 \cdot (1000+111) \\
&= 2^6 \cdot 7 \cdot 11 \cdot 101
\end{align*}

Alternatively, consider adding the first and last terms, and second and second and last terms, etc we obtain $7777$. There are $2 \cdot 4^3$ terms so $7777 \cdot 4^3 = 2^6 \cdot 7 \cdot 11 \cdot 101$