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divyadna
12= 2^2*3^1
No. of factors= (2+1)*(1+1)=6
18= 3^2*2^2
No. Of factors= (2+1)*(2+1)=6
32=2^5
No. Of factors= = 5+1=6

12, 17, and 32 are three numbers less than 40 which have six positive factors.
Answer is option B IMO

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divyadna check 20,28
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How many positive integers less than 40 have six positive factors?

A. 2
B. 3
C. 4
D. 5
E. 6

Solution
Number can be a^p*b^q*c^r
where a, b, c are prime numbers

To calculate the number of factors (p+1)(q+1)(r+1)
to get factor 6
(p+1)(q+1)=6
1*6,2*3,3*2
number can be
=2^5=32
=2^2*3^1=12
=2^1*3^2=18
=2^2*5^1=20
=2^2*7^1=28
Answer is 5
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For an integer to have exactly 6 factors it must either be of the form p^5 or p*q^2 where p and q are primes.

For p^5<40, p can only be 2.
N=32.

For p*q^2<40,(p,q) = (2,3) (3,2) (5,2) (7,2).
So, N=(18,12,20,28).

So in all N has 5 distinct values.

Hence D is correct choice.

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