Are there numbers such that A + B = 10A+B?

Question

I was just wondering, apart from zero,are there numbers where $A+B=10A+B$ (the number AB)?

Answer 1

Hint: Note that $\overline{AB}=10A+B$ ,where $\overline{xy}$ represents the combination of a number. E.G if $x=3$, $y=6$, $\overline{xy}=36$

Solution: \begin{align} A+B&=10A+B \text{Given}\\ 0&=9A \tag{Subtraction Property}\\ 9A&=0 \tag{Symmetric Property}\\ A&=0 \tag{Division Property} \end{align} When $A=0$, your number works. However, if $A=0$, it is not a 2-digit number, and thus does not meet the criteria.

Answer 2

$$A+B=10A+B$$

so

$$A=10A$$....

Answer 3

By $AB$ you apparently mean $A \cdot 10 + B \cdot 1$. Thus you are asking for $A$, $B$ such that $A+B = 10A + B$. This is true only for $A=0$.

Answer 4

If you allow $A=0$, sure. If not, then the sum of two single digit numbers is at most $18$, so $A=1$. That becomes problematic quickly.

Answer 5

$$10A + B + 10C + D = 1000A + 100 B + 10C + D$$ $$10A + B = 1000A + 100 B$$ $$9990A+99B=0$$

Not for $A, B > 0$

Answer 6

We have $$A+B=10A+B⟺A=10A⟺A=0,B∈ℝ$$