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A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors(which are not composite) is 30. How many such numbers are possible?

A. 2
B. 4
C. 6
D. 3
E. Not possible

Firstly , we should note that 1 is NEITHER a prime nor a composite number.The first composite number is 4.Thus, when the problem states that there are 4 factors that are not composite, these nos are 1,2,3,5. Thus, the given number = 2^a*3^b*5^c. Also, (a+1)*(b+1)*(c+1) = 32. We can break down 32 into 3 integers as : 2*2*8 or 4*4*2
Also, the only possible combinations for a,b,c are : 3,3,1 OR 1,1,7. Thus, each combination has 3 possible orders and we have a total of 6 possibilities.

C.
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A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors (which are not composite) is 30. How many such numbers are possible?

A. 2
B. 4
C. 6
D. 3
E. Not possible

Let's see data,
i) a number with 32 factors, out of which 4 are not composite (Too vague data)
ii) product of these 4 factors is 30
lets take (ii) data
30 = 2*3*5*1 (1 is natural number i.e not composite nor prime)
so we know 3 factors of number are 2,3,5
now 32 = 2*4*4, 2*2*8
so total numbers possible = 3C1 + 3C1 = 6
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A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors (which are not composite) is 30. How many such numbers are possible?

A. 2
B. 3
C. 6
D. 8
E. Not possible

There are two types of number when we consider the number of factors they have:-
1) Composite number: More than 2 factors
2) Prime number: Exactly two factors including 1 and itself
number 1 does not fall in any of the two, as it has only one factor itself.

Quote:
A number has exactly 32 factors out of which 4 are not composite
1 is a factor of all numbers, so the 4 factors that are not composite are 1 and 3 prime numbers => \(n=a^xb^yc^z\)
Number of factors = \((x+1)(y+1)(z+1)=32\)

Product of these 4 factors (which are not composite) is 30 : Irrelevant information, although it tells us that the number is \(2^x3^y5^z\).

Now we know that the prime numbers are 3: a, b and c.
cases of 32 as product of three numbers greater than 1.

1) \(32=2*2*8=(1+1)(1+1)(7+1)\)
so x, y and z can be 1, 1 or 7. = \(\frac{3!}{2!}=3\)
\(n=a^1b^1c^7\) or \(n=a^1b^7c^1\) or \(n=a^7b^1c^1\)

2) \(32=2*4*4=(1+1)(3+1)(3+1)\)
so x, y and z can be 1, 3 or 3. = \(\frac{3!}{2!}=3\)
\(n=a^1b^3c^3\) or \(n=a^3b^3c^1\) or \(n=a^3b^1c^3\)

Total = 3+3=6

C
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Given: A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors (which are not composite) is 30.

Asked: How many such numbers are possible?

32 = 2ˆ5 = 2*2*2*4
30 = 1*2*3*5

There are 3 prime factors 2, 3, & 5. No other prime factor is possible.

Therefore, the number is of the form \(2ˆa*3ˆb*5ˆc\)
The number of factors = (a+1)(b+1)(c+1) = 32 = 2*2*8 = 2*4*4

(a,b,c) = {(1,1,7),(1,7,1),(7,1,1),(1,3,3),(3,1,3),(3,1,1)} : Total 6 possibilities

IMO C
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four non composite factors that result in 30 are 1,2,3,5
to find out the total no of factors of the given no it must be in the powers of 2,3,5
since total no of factors of any prime factorization =(a+1)*(b+1)*(c+1)=32 given
so a,b,c can be either of the form 1,1,7 or 7,1,1 or 1,7,1 OR 1,3,3 or 3,1,3 or 3,3,1
total of six different numbers possible.
OR alternately 1,1,7 and 1,3,3 can be distributed among the powers of 2,3,5 in 3!/!2=3 ways each
i.e. 3+3=6 different permutations are possible resulting in 06 different numbers.
so C
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