Discover the Pattern of PDS Numbers and Find the Nth PDS Number

[shekhar kumar]

PDS number:> a number whose product of digits is completely divisible by sum of its digits.
Now the series comes out to be: 1,2,3,4,5,6,7,8,9,10,20,22,30,36,40,44,50,60,63,66,70... so on.
(1. Numbers from 1 to 10 are already divisible by themselves.
2. Then all the multiples of 22 follow the pds series completely.
3. And finally all the numbers which contains atleast one zero are also pds number).
I have to find the general term of this pds series so that i can find the nth pds number in a single shot.
Plz help in doing that as soon as possible.
Thanks in advance...

 

[karthikccet]

you have said three conditions...in that case how the numbers 36,63...will come in the list...they are neither containing zero...nor multiples of 22...

PDS number:> a number whose product of digits is completely divisible by sum of its digits.
Now the series comes out to be: 1,2,3,4,5,6,7,8,9,10,20,22,30,36,40,44,50,60,63,66,70... so on.
(1. Numbers from 1 to 10 are already divisible by themselves.
2. Then all the multiples of 22 follow the pds series completely.
3. And finally all the numbers which contains atleast one zero are also pds number).
I have to find the general term of this pds series so that i can find the nth pds number in a single shot.
Plz help in doing that as soon as possible.
Thanks in advance...

 

[Bipolarity]

PDS number:> a number whose product of digits is completely divisible by sum of its digits.
Now the series comes out to be: 1,2,3,4,5,6,7,8,9,10,20,22,30,36,40,44,50,60,63,66,70... so on.
(1. Numbers from 1 to 10 are already divisible by themselves.
2. Then all the multiples of 22 follow the pds series completely.
3. And finally all the numbers which contains atleast one zero are also pds number).
I have to find the general term of this pds series so that i can find the nth pds number in a single shot.
Plz help in doing that as soon as possible.
Thanks in advance...

I don't think there is any general formula for finding the nth term. You could of course write an algorithm that generates the nth digit but the time would increase exponentially with the size of n.

BiP