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.
 A: NOTE: For your work, we also have cases $(2,9)$, $(3,8)$ ... to check.
Here's what I did:
We start from the left, by the fact that $7 + 4 = 1a$, we must have $a = 1$ or $a = 2$, since either there is a carry or not.  
From the one's digit, we have that :
For $a = 1$, we must have $b = 0$.
For $a = 2$, we must have $b = 9$.
Now let us try it, we have $710 + 401 = 1111$, which is impossible.
We have $729 + 492 = 1221$, which corresponds to a solution.
A: $ab+ba$ is either $21$ or $121$. Well, it can't be $21$, so it's $121$. That means that we carry 1 over to the hundreds' place.
Looking at the hundreds and thousands we have $7+4+1=1a$, and from there it's easy.
A: You can also bring in the divisibility test for $11$ into the problem.  First note that $700+400=1100$ and $ab+ba=11(a+b)$, forcing your sum $1a21$ to be a multiple of $11$.
By the divisibility test for $11$, then, $1-a+2-1$ must be a multiple if $11$, forcing $a=2$.  The ones digit sum then must be $b+2=11(\ge 2)$, thus $b=9$.
A: One example is 729+492=1221
a=2,b=9
   7 a b
 +4 b a
1 a 2 1
Hint: b+a is not equal to 1. So b+a=11 hence the solution
A: 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
$$
\boxed{7\underline{29} + 4\underline{92} = 1\underline{2}21}.
$$
