A two-digit number $A$ is a supernumber if it is possible to find two numbers $B$ and $C$, both also two-digit numbers, such that:

- $A = B + C$;
- the sum of the digits of $A$ = (sum of the digits of $B$) + (sum of the digits of $C$).

For example, 35 is a supernumber. Two different ways to show this are $35 = 11 + 24$ and $35 = 21 + 14$, because $3 + 5 = (1 + 1) + (2 + 4)$ and $3 + 5 = (2 + 1) + (1 + 4)$. The only way to show that 21 is a supernumber is $21 = 10 + 11$.

a) Show in two different ways that 22 is a supernumber and in three different ways that 25 is a supernumber.

b) In how many different ways is it possible to show that 49 is a supernumber?

c) How many supernumbers exist?
Solution

a) Two ways to show that 22 is a super number are $22=10+12$ and $22=11+11$, since $2+2=$ $(1+0)+(1+2)$ and $2+2=(1+1)+(1+1)$.

b) We present below all the ways to write 49 as the sum of two two-digit numbers, always placing the smaller of the two on the left:

$$
\begin{aligned}
49 & =10+39 \\
49 & =11+38 \\
49 & =12+37 \\
& \vdots \\
49 & =23+26 \\
49 & =24+25
\end{aligned}
$$

A simple count reveals that the number of ways is $15=24-10+1$. Note that any of these can be used to show that 49 is a super number. For example, in the first one, we have that $4+9=(1+0)+(3+9)$. Therefore, it is possible to show that 49 is a super number in 15 different ways. c) Since 10 is the smallest two-digit number, $20=10+10$ is the smallest two-digit number that can be written as the sum of two other two-digit numbers. Now consider any two-digit number $x$ that is greater than or equal to 20, and let $a$ be its tens digit and $b$ be its units digit. Let's now think about the number $x-10$. This is a number greater than or equal to 10, since $x$ is greater than or equal to 20. Therefore, it has two digits. Thus, it can be written as $mn$, where $m$ is the tens digit and $n$ is the units digit. The tens digit is $m=a-1$ and the units digit is $n=b$. Now, writing $x=10+(x-10)$, we see that $x$ is a super number, since

$$
\underbrace{a+b}_{x}=\underbrace{(1+0)}_{10}+\underbrace{(a-1)+b}_{x-10}
$$

An example helps to understand this reasoning. Consider $x=38$; here we have

$$
\begin{array}{r}
38 \\
-10 \\
\hline 28
\end{array}
$$

that is, $x-10=28$. The expression $x=10+(x-10)$, in this case, is $38=28+10$, which shows that 28 is a super number, since

$$
3+8=11=(1+0)+(2+8)
$$

Therefore, all numbers from 20 to 99 are super numbers, and they are in number $99-20+1=80$.