Just giving a try

IMO C.

1. N=2^5(k) where k = 1,3,5,7,9,11,13,..... (Not multiple of 2 as 2^6 will not divide N)
Not Suff as u cannot say anything about R

2. N=5^6(k') where k' = 1,2,3,4,6,7,8,9,11,12,13,14...... (Not multiple of 5 as 5^7 will not divide N)
Not Suff as u cannot say anything about R

Now, combine both then N= (2^5) * (5^6) (k") where k" = 1,3,7,9,11,13,17.... (Not multiple of 2 and 5 as 10^(R+1) will not divide N)

N= (10^5) * 5 * k" and as we will not encounter any more 2 and 5 in k",R has to be 5.

Hope I am correct. Let me know if it helps or other suggestions are always welcome.

shovan85 wrote:Just giving a try

IMO C.

1. N=2^5(k) where k = 1,3,5,7,9,11,13,..... (Not multiple of 2 as 2^6 will not divide N)
Not Suff as u cannot say anything about R

2. N=5^6(k') where k' = 1,2,3,4,6,7,8,9,11,12,13,14...... (Not multiple of 5 as 5^7 will not divide N)
Not Suff as u cannot say anything about R

Now, combine both then N= (2^5) * (5^6) (k") where k" = 1,3,7,9,11,13,17.... (Not multiple of 2 and 5 as 10^(R+1) will not divide N)

N= (10^5) * 5 * k" and as we will not encounter any more 2 and 5 in k",R has to be 5.

Hope I am correct. Let me know if it helps or other suggestions are always welcome.

many thanks for your explanation!!!!

alivapriyada wrote:
many thanks for your explanation!!!!

You are welcome

alivapriyada wrote:N and R are integers. If N is divisible by 10^R, but cannot be divisible by 10^(R+1), R=?

1). N is divisible by 2^5, but cannot be divisible by 2^6.
2). N is divisible by 5^6, but cannot be divisible by 5^7.

[spoiler]OA not known, plz explain!!![/spoiler]

tobe honest, i dont take care of data these days, i only do PS and today when i take a look at this question, i guess it might be C but to tell me to solve it in detail i cant....and lucky me the answer is C
and i think i should review the way i study...