Find digits $a,b$ such that $7ab + 4ba = 1a21$.

Question

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.

Answer 1

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.

Answer 2

$ab+ba$ 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.

Answer 3

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

Answer 4

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 $$ \boxed{7\underline{29} + 4\underline{92} = 1\underline{2}21}. $$

Answer 5

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 $b=9$.