1nope[1]
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..can be..can be (ans batao)[3]
btw this is a q of bmat open test of last year
-
UP 0 DOWN 0 0 6
6 Answers
1105=3*5*7=(2+1)*(4+1)*(6+1)
so N=a2*b4*c6
where a,b and c are prime numbers
as 1000 is a factor so one them will be 2 and the other will be 5
and a is to be determined
we can proceed in this way and take the various cases
621000 = 23.53
also n is a factor of 1000
now since n has 105 factors, and 105=5x3x7
it means that n can have atmost 3 prime factors of n.. 2 of them being 2 and 5
Assume that there are exactly 3 factors
This means that n is of the form 2a 5b pc
the number of factors being (a+1)(b+1)(c+1) = 5x3x7
so a, b and c are either 4 or 2 or 6
but a is greater than equal to 3 .. so a+1>3...
also b+1 is greater than 3 by the same logic
The number of odd factors will be (b+1)(c+1)
so it will be a factor of 105 which can take values either 3x7 or 3x5 = 21 or 15
Assume that there are exactly 2 factors
This means that n is of the form 2a 5b
the number of factors being (a+1)(b+1) = 5x21 = 3x35=7x15
again we know that a is greater than equal to three
The number of odd factors will be (b+1) = 21 or 5 or 7 or 15
So the answer is 4 values given by 4, 6, 14 and 20.
I hope i have not missed out anything [1]