CAT 2005 | Question: 19

Question

For a positive integer $n$, let $\text{P}_n$ denote product of the digits of $n$ and $\text{S}_n$ denote the sum of the digits of $n$ The number of integers between $10$ and $1000$ for which $\text{P}_n + \text{S}_n = n$ is

• A. $81$
• B. $16$
• C. $18$
• D. $9$

Answer 1

n can be a 2 digit or a 3 digit number.

Case (1) Let n be a 2 digit number.

Let n = 10x + y

$P_{n}$  = xy and $S_{n}$ = x + y

Now, $P_{n}$ + $S_{n}$ = n

 xy + x + y = 10x + y

xy = 9x = 9

There are 9 two digit numbers (19, 29, 29, … ,99) for which y = 9

Case (2) Let n be a 3 digit number.

Let n = 100x + 10y + z

$P_{n}$ = xyz and $S_{n}$ = x + y + z

Now, $P_{n}$ +$S_{n}$ = n xyz + x + y + z = 100x + 10y + z 

xyz = 99x + 9y

 z = $\frac{99}{y} + \frac{9}{x}$  

From the above expression, 0 < x, y < 9

But, we cannot find any value of x and y, for which z is a single digit number.

There are no 3 digit numbers which satisfy$P_{n}$ +$S_{n}$ = n

Hence, option (4)9 is the Answer.