# $pqrs \cdot 4 =srqp$,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.

• 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 $r$).

• Why is it when the RHS<8000,s=3 will be rejected? Commented Jun 25, 2016 at 13:55
• Ah,okay I see,then at the last part where,$13q-2r+1 =0$,we don't know the values of $q$ & $r$,so we use 'trial and error' to find the suitable values? Commented Jun 25, 2016 at 14:01
• Note that $s \ge 8$ and you don't need to check $s=3$! Commented Jun 25, 2016 at 14:02
• @Arc Noepi That's right. Actually, if $q\ge 2$ then even for $r = 9$ the LHS of the equation will be always positive. Thus, $q = 1$ and immediately we can find $r$. Commented Jun 25, 2016 at 14:03
• I see!Okay,I understand this way of solving already!But if I use the method I used above,I would not be able to solve the question if I don't deduce that $s=3$ or $8$ right? Commented Jun 25, 2016 at 14:07
• 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 $qrs=\boxed{178}$.
• How can we infer that $s=8$ or$s=9$, I don't really understand how $s=9$? Commented Jun 25, 2016 at 14:29
• Because $pqrs=2qrs\geq 2000$, we have $srpq=4\times pqrs\geq 8000$ and so $s\geq 8$. So it must be that $s=8$ or $s=9$. And I've already explained how to eliminate the case $s=9$. Commented Jun 25, 2016 at 14:35
• Oh,Okay I understand, thanks Commented Jun 25, 2016 at 14:40

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 $qrs=178.$

• Wow,this is really a nice method!I understand this the easiest. Commented Jun 25, 2016 at 16:19

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 $r=7$

• I understand the first paragraph,but I'm lost at the part where $2*4 =8$,so $s=8$ Commented Jun 25, 2016 at 13:58
• How do we know $q=1$ or $q=2$?
– Ovi
Commented Jun 25, 2016 at 13:59
• because if we set $q=3$,then $3*4=12$,and $2*4+1=9$ not 8 Commented Jun 25, 2016 at 15:22