<s>[INST] Find digits $a,b$ such that $7ab + 4ba = 1a21$. I have to find all the digits $a$ and $b$ such that $7ab + 4ba = 1a21$.
Note: there is no multiplication, those are three decimal numbers.
I put this equality this way: $$7\cdot100 + a \cdot 10 + b + 4 \cdot 100 + b \cdot 10 + a = 1 \cdot 1000 + a \cdot 100 + 2 \cdot 10 + 1$$
$$1\cdot 1000 + (a-11)\cdot 100+(2-a-b)+1-a-b=0$$
So, our number will be something like this: $1(a-11)(2-a-b)(1-a-b)=0$. This is contradiction since the first digit is $1$, and on the right we have $0$.
Does that mean there are no $a$ and $b$ to satisfy this equality? 
Alternatively, I did the following: since the last digit in $1a21$ is $1$, it has to be $a+b=L1$. $a$ and $b$ are digits, so there are couple of cases: $a=0$ and $b=1$, $a=1$ and $b=0$, $a=5$ and $b=6$, $a=6$ and $b=5$... None of those cases satisfy what we want, so there are no $a$ and $b$.
Are those two ways to do it the right ones? I don't have any solutions therefore I'm not sure.
Thank you.
 [/INST] Another approach to solve such equations is to write it as an algebraic equation, instead of with base-10 digits. We get
$$
700 + 10a + b + 400 + 10b + a = 1000 + 100a + 21
$$
which simplifies to:
$$
11(a + b) + 1100 = 1021 + 100a
$$
or
$$
11b + 79 = 89a.
$$
Now we use the fact that $a$ and $b$ are between $0$ and $9$. In particular, $11b$ is at most $99$, so $11b + 79$ is at most $99 + 79 = 178$, which equal to two times $89$. So $a$ can be at most $2$.


*

*If $a = 2$, then $11b + 79 = 178$, so $b = 9$.

*If $a = 1$, then $11b = 10$, so this doesn't work.

*Finally, if $a = 0$, then $11b + 79 = 0$, so this doesn't work either.
Thus the only solution is $a = 2, b = 9$. We have
$$
\boxed{7\underline{29} + 4\underline{92} = 1\underline{2}21}.
$$
</s>