This is question 26 from Australian Maths Competition 2013.
$pqrs $ is a 4-digit number and has the property that $pqrs \cdot 4 = srqp$.If p=2,what's the value if the 3-digit number qrs?
Here's what I tried.
$(2000 + 100q+10r+s)(4)=1000s+100r+10q+2$
So,
$ \frac {1000s+100r+10q+2}{4} $= some integer.
$250s+25r+ \frac {5q}{2} + \frac 12 =2000+100q+10r+s$
$ \frac {5q}{2} $ should be an fraction so that when it adds up with half, it will be an integer. So,$q$ must be an odd number.
$pqrs \cdot 4 =srqp $,where $p=2$
$4s= 10j +2$, where j is some integer
$250s=625j + 125$
And then I'm lost.
Best Answer
Firstly, we notice that it must hold that $4s$ must end at the digit $2$. Thus, there are two options for $s$. Either $s = 3$ or $s = 8$.
1st case: $ s = 3$. In that case we have: $$8000 + 400q + 40r + 12 = 3000 + 100 r +10 q + 2.$$ But in that case the RHS is always less than $8000$, thus $s = 3$ is rejected. Hence, $\boxed{s=8}.$ In that case, we have:
$$\begin{array}[t]{l}8000 +400q + 40 r + 32 = 8002 +100r +10q\\ 390q - 60r +30 = 0\\ 13q - 2r +1 =0.\\ \end{array}$$
Testing the values for $q,r $ we have that $\boxed{r = 7}$ and $\boxed{q=1}$ (actually $q$ cannot be greater than $1$, because the above equality will always fail for every possible value of $r$).