Find replacements for each letter to make the problems true. For each problem the letter represents a different single digit and must be the same wherever it appears

A B C D x 4 = D C B A

and

R S S + S T = S R R

There are a couple of quite different methods suggested here, so if you don't like the first one, do read on to Arvan's message later.

$100R+10S+S+10S+T=100S+10R+R$

$\Rightarrow 89R-79S+T=0$

since $R$, $S$ and $T$ are each digits we know

$0\le R\le 9$, $0\le S\le 9$, $0\le T\le 9$

So now try whole numbers in the equation above (try values of $R$ and $S$ and check that $T$ is then between 0 and 9). When you have your values check them in the original equation.

You can use the same method for the first question (I think). Jim

For the first question I would leave the C's on the right hand side of the equation and other letters on the left. Then you can try values of A, B and D that will give a whole number for C (try dividing both sides of the equation by 60 before you put numbers in: this should make sense when you get there).
I should point out that A, D, R and S can't be 0 because they are the first digit in the numbers.
This method works, but not quite as well for the first question, anyone else have any ideas for solving this problem?

Jim

Wow!
I think I can do it now.

Thank you Jim

I think there is a solution to 1, here is what I got. There may be other solutions, I don't know.
If you don't want to see the answers, please look away now :)

1. $A=2$, $B=1$, $C=7$, $D=8$

   $ABCD\times 4=DCBA$

   $2178\times 4=8712$

2. $R=7$, $S=8$, $T=9$

   $RSS+ST=SRR$

   $788+89=877$

1. $3999A+390B-996D=60C$ (then $÷60$)
2. $89R-79S+T=0$ (I suppose $79S-89R=T$ is more useful).

I usually try to work out restrictions for each digit rather than turning the whole thing into an equation:

For 1:

So $A$ (units of $DCBA$) is even and (thousands of $ABCD$) less than 3, so $A=2$.

Hence $D=8$, now we can put these in and simplify to

$BC\times 4+3=CB$

So $B$ is odd (units of $CB$), less than 3 (tens of $BC$) so $B$ is 1.

Hence $C$ is 2 or 7 (units of $4C+3$ are 1) and at least 4 (tens of $CB$)

So the solution is $2187\times 4=8712$.

That's a much better method Arvan, I like it.

Jim

This method is perhaps even more useful on the second question: write the sum in columns, and then compare the tens and the units columns. What does this tell you?

There are lots more alphanumeric problems on the NRICH site. Try these links:
Long multiplication
Skeleton (long division)
Kids (addition)
Tis Unique (addition)
Cayley (addition)
Aba (addition)
Two and two (addition)
Alphabet soup (inventing your own)
ABC (long multiplication)
Football sum (addition)
Star in the East (addition)
Many of these will respond better to Arvan's method than Jim's.