for two digit integers, \(n=ab\) with digits a and b

\(P_n = a*b\)
\(S_n = a+b\)
and \(n = ab = 10a + b\)

\(P_n + S_n = n\)
i.e. \(a*b + (a+b) = 10a+b\)
i.e. \(a*b = 9a\)
i.e. \(b = 9\) for ny value of a

so total two digit number \(n = ab = {19, 29, 39, 49, 59, 69, 79, 89, 99} = 9\) Numbers


for THREE digit integers, \(n=abc\) with digits a and b

\(P_n = a*b*c\)
\(S_n = a+b+c\)
and \(n = ab = 100a + 10b + c\)

\(P_n + S_n = n\)
i.e. \(a*b*c + (a+b+c) = 100a+10b+c\)
i.e. \(a*b*c = 99a+9b = 9(11a+b)\)
if we take a = 1 , and b = 1 then, c = 99 + 9 = 108
but a ,b , c should be 1 to 9 .
hence, there is no solution from here .

so total number \(n = ab = {19, 29, 39, 49, 59, 69, 79, 89, 99} = 9\) Numbers

ANswer:OPtion E
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So, the number n can be 2-digit or 3-digit.
(I) Say n is 2-digit : xy = 10x+y
Now \(P_n=x*y\) , and \(S_n=x+y\)
\(P_n+S_n=x*y+x+y=10x+y........x*y=9x.....y=9\)
So x can be anything from 1 to 9, and y is 9. Hence 9*1=9 numbers => 19, 29, 39...89, 99

(II) Say n is 3-digit : xy = 100x+10y+z
Now \(P_n=x*y*z\) , and \(S_n=x+y+z\)
\(P_n+S_n=x*y*z+x+y+z=100x+10y+9........x*y*z=99x+9y.....\)
Now, even a portion of 99x+9y, \(99x=9*11*x\neq{x*y*z}\) as 11 is not a single digit.
No values possible

Total = 9

E