Let N has total 105 factors including 1 and N. If N is divisible by 1000 then total number of odd factors of N lying between 1 and N can be
1. 4
2. 6
3. 14
4. 20
621000=2353
if the number is 1000. K
then it is obvious that the third prime other than 2 and 5 will have power 2. (even)
N has in all 105 factors
105=3.5.7
thus the powers are 2, 4, 6
105=15.7.1
thus the powers are 14, 6
105=21.5.1
thus the powers are 20,4
2 and 5 will have either power 4 or power
hence the no of odd factors will be 5.3 or 7.3 ie 15 or 21
but if lying between 1 and N then it will be 14 or 20!
and from the above the possibility of 14.1 and 20 is also there.... check that.. and hence the answer is all the above...
62because 3.5.7=(1+2)(4+1)(6+1)
so powers of 3 factors will be 2, 4, or 6
One of these will be the power of 2 .. but 23 is a factor of N.
Hence 2 cannot be the power of 2
only 4 or 6 can be.
hence the answer...
11Ans.1,2,3,4
105 = 3 × 5 × 7 and 1000 = 23 × 53.
It means N has at least two prime factor 2 or 5 each having at least 3 powers.
It means powers of 2 or 5 in N is either 5 - 1, 7 - 1, 3 × 5 - 1, or 3 × 7 - 1
Total number of odd factors is either 7 × 3 - 1, 3 × 5 -1, 7 - 1 or 5 - 1
Total number of odd factors is either 20, 14 , 6 or 4
this was the solution
@unique i write them online
62yes subash you are right i made a mistake in the answer :(
The third power has still to be even..
Question: What is the power of a perfect square? even or odd...
I am correcting it :)
