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03 Jul 2025, 06:39
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Minimum number of prime factors is 1: example: 7^23
24=3*2^3
As each distinct prime number contributes, at least, multiplying by 2 the number of positive factors, to maximize the number of prime factors I choose a number like this: 2*3*5*7^2, that have 4 distinct prime factors.
The range between 1 and 4 is 3.
Correct answer is C
24=3*2^3
As each distinct prime number contributes, at least, multiplying by 2 the number of positive factors, to maximize the number of prime factors I choose a number like this: 2*3*5*7^2, that have 4 distinct prime factors.
The range between 1 and 4 is 3.
Correct answer is C
03 Jul 2025, 06:56
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Bunuel
Formula used:
Number of factors of a number:
If N= a^p*b^m*c^q
The number of factors of N: (p+1)*(m+1)*(q+1)
Minimum number of prime factors: N=a^23
=> 23+1=24 factors
Here, there is 1 distinct factor
Maximum number of distinct prime factors:
a^1*b^1*c^1*d^1
This gives:
(1+1)*(1+1)*(1+1)*(1+1)=16
We cannot have greater than 4 distinct factors because: 2^5=32 >24(factors given in question)
Here, there are 4 distinct factors
=> Range: 4-1=3
Answer: C
03 Jul 2025, 07:09
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The minimum distinct prime factor n can have is 1 when the n is of the form - (prime no.)^23
On the other hand, the maximum distinct prime factors can be found by expanding 24 - 3* 2*2*2
It follows that n can be of the form - (pr.no 1)^2* (pr.no 2)^(pr.no 3)*(pr.no 4) in order to get a total of 24 positive factors. Maximum therefore is 4.
Range between the min and max is 3.
Therefore, Option C
On the other hand, the maximum distinct prime factors can be found by expanding 24 - 3* 2*2*2
It follows that n can be of the form - (pr.no 1)^2* (pr.no 2)^(pr.no 3)*(pr.no 4) in order to get a total of 24 positive factors. Maximum therefore is 4.
Range between the min and max is 3.
Therefore, Option C
