$pqrs \cdot 4 =srqp $,then what is the value of $qrs$? 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.
 A: *

*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 $r$).
A: *

*With $4\times pqrs=srqp$ and $pqrs\geq 2000$, we infer that $s=8$ or $s=9$. Together with the fact that $\times pqrs$ ends in $p=2$, it must be that $s=8$.

*So we have
$$
4\times(2000+100q+10r+8)=8000+100r+10q+2\implies 13q=2r-1.
$$
Possible values for $2r-1$ are $\{-1,1,3,5,7,9,11,13,15,17\}$. It's obvious then that we must have $q=1$ and $r=7$.

*So, overall, the answer is $qrs=\boxed{178}$.

A: Here is yet another way to look at it:
$$
8000+400q+40r+4s=1000s+100r+10q+2 \iff 7998+390q=60r+996s.
$$
Since the number on the LHS must end in an 8, the number on the RHS must also end in 8. This means that either $s=3$ or $s=8$. If $s=3$ then
$$
7998+390q=60r+2888 \iff 5010+390q=60r \iff 501+39q=6r. 
$$
Since the smallest the LHS can be is 501 and the largest the RHS can be is 54, we conclude there is no solution. So we reject $s=3$ and let $s=8$ so that
$$
7998+390q=60r+7968 \iff 30+390q=60r \iff 3+39q=6r. 
$$
Now the LHS has to be between 0 and 54 (because $r\in\{0,\ldots,9\}$) which means $q=0$ or $q=1$. If $q=0$ then we have 
$$
3=6r
$$
which has no integer solutions, so we must have $q=1$. Thus
$$
42=6r \iff r=7.
$$
Thus $qrs=178.$
A: I think you should analysis this question.
At first,$$2qrs*4=srq2$$
and we know,$8*4=32$,and $ 3*4=12$
and $2*4=8$,so   $s=8$
and because $s=8$,so $q=1,or,2$,
if we set $q=2$ and $4r+3=2$ is not true,so q=1
and next,we know $r=7$
