Question

What is the sum of all natural numbers $n$ such that the product of the digits of $n$ (in base 10) is equal to $n^2-10n-36$ ?

• A. 12
• B. 13
• C. 124
• D. 2612

Answer 1

The correct option is B 13

Product of two natural number will be non negative, so
$n^2-10n-36\ge 0\Rightarrow n\in (\infty ,5-\sqrt{61}]\cup [5+\sqrt{61},\infty )$
As $n\in N$ so $n\ge 13$,

Case 1 : 2 digit natural number $n$
The maximum value of the product of the digits $=9\times 9=81$
$\therefore n^2-10n-36\le 81$
$\Rightarrow n\in [5-\sqrt{142},5+\sqrt{142}]$
As $n\in N$ so $13\le n<17$,
(i) For $n=13$ product of digits $=3$
$169-130-36=3$
Hence it satisfies the relation given.
(ii) For $n=14$ product of digits $=4$
$196-140-36\ne 4$
relation not satisfied.
(iii) For $n=15$ product of digits $=5$
$225-150-36\ne 5$
relation not satisfied.
(iv) For $n=16$ product of digits $=6$
$256-160-36\ne 6$
relation not satisfied.

Case 2 : 3 digit natural number $n$
The maximum value of product of digits $=9\times 9\times 9=729$
The smallest 3 digit natural number can be $=100$
$\therefore {100}^2-100\times 10-36=8964>729$
So there is no 3 digit number possible satisfying the relation.

Hence $n=13$