Basic Number Theory

Python:

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for n in xrange(10000,100000):
    s = str(n)
    m = int(s[0:2]+s[3:5])
    if n % m == 0:
        print "Answer:", n
        break

Let $n=10000a+1000b+100c+10d+e$ and $m=1000a+100b+10d+e$ , where $a, b, c, d,$ and $e$ are base-10 digits and $a \neq 0$ . If $\frac{n}{m}$ is an integer, then $m|n$ , or

$1000a+100b+10d+e|10000a+1000b+100c+10d+e$ .

This implies that

$1000a+100b+10d+e|9000a+900b+100c$ .

Clearly we have that $8(1000a+100b+10d+e) \\<9000a+900b+100c \\<10(1000a+100b+10d+e)$ , as $a \geq p$ . therefore $\frac{n}{m}$ must be equal to $9$ , and

$9000a+900b+90d+9e=9000a+900b+100c$ .

This simplifies to $90d+9e=100c$ . The only way that this could happen is that $c=0$ . Then $d=e=0$ . Therefore the only values of $n$ such that $\frac{n}{m}$ is an integer are multiples of $1000$ . And therefore the least value is $10000$ .