a) Find all positive two-digit integers $\overline{a b}$ such that:

$$
(a+1)(b+1)=\overline{a b}+1
$$

b) Find all positive three-digit integers $\overline{a b c}$ such that:

$$
(a+1)(b+1)(c+1)=\overline{a b c}+1
$$
Solution

a) We need to find all numbers of the form $\overline{a b}$ such that

$$
(a+1)(b+1)=\overline{a b}+1
$$

We notice now that $\overline{a b}=10 a+b$. Substituting this into the equation above, we get:

$$
(a+1)(b+1)=10 a+b+1
$$

Simplifying this equation, we obtain

$$
a b=9 a
$$

Since the number $\overline{a b}$ has two digits, we have that $a \neq 0$. Thus, we can divide both sides of the equation above by $a$ to get $b=9$. Therefore, the possible values for $\overline{a b}$ are:

$$
19, \quad 29, \quad 39, \quad 49, \quad 59, \quad 69, \quad 79, \quad 89 \quad \text { and } 99
$$

It is simple to verify that all these numbers satisfy the condition of the problem. For example, for the number 19, we have $a=1$ and $b=9$, so $(a+1)(b+1)=20=19+1=\overline{a b}+1$. For the other numbers, we leave the verification to the reader!

b) We now need to find all numbers of the form $\overline{a b c}$ such that

$$
(a+1)(b+1)(c+1)=\overline{a b c}+1
$$

Note that $\overline{a b c}=100 a+10 b+c$. Thus, the expression above can be rewritten as:

$$
(a+1)(b+1)(c+1)=100 a+10 b+c+1
$$

The left side can be written as:

$(a+1)(b+1)(c+1)=a(b+1)(c+1)+(b+1)(c+1)=a(b+1)(c+1)+b(c+1)+c+1$.

Substituting this into the previous expression and simplifying, we then get:

$$
a(b+1)(c+1)+b(c+1)=100 a+10 b
$$

Now, rearranging the terms, we obtain the following equivalent equation:

$$
0=a\{100-(b+1)(c+1)\}+b\{10-(c+1)\}
$$

However, $b$ and $c$ denote digits, so as numbers they take values less than or equal to 9. Therefore, $(b+1)(c+1) \leq 100$. It follows that $\{100-(b+1)(c+1)\} \geq 0$. Similarly, we observe that $\{10-(c+1)\} \geq 0$. Thus, for the right-hand term in equation (.6) to be equal to zero, it must be that

$$
a\{100-(b+1)(c+1)\}=0 \quad \text { and } \quad b\{10-(c+1)\}=0
$$

Since $a \neq 0$, then $\{100-(b+1)(c+1)\}=0$. This is only possible if $b=c=9$. We conclude that the only possible numbers are

$$
199, \quad 299, \quad 399, \quad 499, \quad 599, \quad 699, \quad 799, \quad 899 \text { and } 999
$$

Now, it is simple to verify that all these numbers satisfy the condition of the problem.