I'm having difficulty in assess this problem in a method without relying too much in too many trial and error. I have noticed there is a similar puzzle but in such it includes the value of one of the digits, but how do you solve this without knowing any other information?.
The problem is as follows:
Using the expression from below:
$$\frac{\overline{ASIR}}{\overline{RISA}}=4$$
Find $\overline{RI}-\overline{SA}$
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\textrm{75}\\ 2.&\textrm{-75}\\ 3.&\textrm{57}\\ 4.&\textrm{-57}\\ 5.&\textrm{-48}\\ \end{array}$
How exactly should this problem be assesed without just guessing?. Can someone help me with this?.
So far the only thing which I could come up with was with the following:
$\overline{ASIR}=4\overline{RISA}$
$1000A+100S+10I+R=4000R+400I+40S+4A$
From this it can be obtained:
$996A+60S=3996R+390I$
But that's it. That's the place where I came stuck. From then on I think its just trial and error, but on looking on the equation all letters with the exception of R and A can take values between 1 to 9. So how to constrain those or to solve this thing in a more orderly manner?. Can someone help me with a step-by-step approach and the most detailed as possible?.
Write it as $\overline {ASIR} \times 4 = \overline {RISA}$. Note that $A$ can only be $1$ or $2$ because otherwise multiplying by $4$ would carry. It has to be $2$ because it is the ones digit of the product. Now $R$ must be $3$ or $8$ to make the $A$ in the product, and it must be $8$ because it is at least $4$. We are left with $\overline {SI} \times 4 +3=\overline {IS}$, so $S$ is odd and must be $1$ to avoid a carry. Then $I$ must be $7$ and we have $2178 \times 4 = 8712$