N is a natural number between 10 and 1000. P denotes the product of its digits. S denotes the sun of its digits. If 6P + 4S= 4N, how many values can N take?

Hi Shubham,

Let the number be abc

Product of the digits = a*b*c
Sum = a+b+c
6*a*b*c + 4*(a+b+c) = 4*(100a + 10b + c)
=> 6*a*b*c + 4a + 4b + 4c = 400a + 40b + 4c
=> 6*a*b*c = 396a + 36b
=> a*b*c = 66a + 6b
=> a*(b*c - 66) = 6b

Now, b is a positive integer. So, RHS is positive => LHS should also be positive => b*c > 66
The only possibilities are b = 8 and c = 9 or b = 9 and c = 8
In both the cases, b*c = 72 => LHS = 6a and RHS = 6b => a = b

So, the two numbers that satisfy the condition are 889 and 998

Aditya did it nicely for 3 Digits for 2 digits, refer below

Let XY be the numbe r
P= X*Y and S=X+Y
From given equation, we have
6P + 4S = 4N
6X*Y + 4X + 4Y = 4(10X + Y)
or, 6X*Y = 36X
or, Y=6
so 16, 26, 36,.............96 counts to 9.
plus 2 more
total 11.

Thank you so much for the explanation about three digit number. I got that now please tell me if the number is 2digit what are the possibilities?

Still i am confused that number can be in a form of (10a+b) also right?