Problem 6

(Problem posed by Department of Mathematics, Sarada Vilas College)

It is well know that there are two digit natural numbers \(ab\) (with \(a\neq0\)) such that

\((a+b)+(a\times b) = ab\),

where \(ab\) is not the product of \(a\) and \(b\). It is the two digit number \(ab\) just like \(12\). That is, on adding the sum of its digits to the product of its digits, we get back the same two digit number. For instance,

\((6+9)+(6\times 9) = 69\).

It is not difficult to see that there are more such numbers. Solve the following two problems:

(1) Find all two digit natural numbers \(ab\) (\(0 < a \leq 9 \), \(0\leq b \leq 9\)) such that

\((a+b) + (a\times b) = ab \)

(2) Find all three digit natural numbers \(abc\) (\(0 < a \leq 9, 0\leq b \leq 9\) and \(0\leq c\leq 9\)) such that

\((a+b+c)+(a\times b \times c)\) = abc

Remainder. \(abc\) doesn't mean the product \(a\), \(b\), and \(c\). It is the three digit number \(abc\), like \(123\).

Without justifications, your answers would be incomplete.

(Comment the values in the comment box and send the detailed solution to dep.math.svc@gmail.com on or before 27th September, 2023.)