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VeritasKarishma, chetansharma, generis, Bunuel
Can you please explain the solution in layman's terms, if possible. After going through the solution multiple times, I am unable to make heads or tails out of it.
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Bunuel
A positive integer is called semiprime if it is the product of exactly two not-necessarily-distinct prime numbers. A positive integer is called highly composite if it has more factors than any smaller positive integer has. How many positive integers are both semiprime and highly composite?

A. 0
B. 1
C. 2
D. 3
E. Infinitely many

We are dealing with semi-prime and highly composite.
Let us find out the number of factors of semi-prime :
1) Both the prime are same - \(a^2\)......Number of factors = \((1+2)=3\)
2) Both the prime are different - \(a*b\)......Number of factors = \((1+1)(1+1)=2*2=4\)

Now let us check for the first number where the factors become equal to more than 4.
\(a^3, a^2b\), are few of the least numbers.
Lowest number : \(a=2...a^3=8\)...8 is not semi prime and has factors equal to 4. So, the semi-primes satisfying the given conditions have to be less than 8.
Semi-prime less than 8 are 2*2, 2*3. 2*5 or 3*3 become more than 8.
Answer - 4 and 6 are the semi-primes which are also highly composite.

C
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Bunuel
A positive integer is called semiprime if it is the product of exactly two not-necessarily-distinct prime numbers. A positive integer is called highly composite if it has more factors than any smaller positive integer has. How many positive integers are both semiprime and highly composite?

A. 0
B. 1
C. 2
D. 3
E. Infinitely many


The question is based on factorisation.

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/1 ... ly-number/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/1 ... t-squares/

Semi-prime numbers have exactly 2 prime factors - same or different.

So a semi prime number N, would be a^2 or ab where a and b are both primes.
So N would have 3 or 4 factors. For it to also be highly composite, all positive integers (starting from 1) less than N should have 3 or 4 factors. The moment you get 5 factors or more, you cannot have a number which is both semi prime and highly composite.
The first number to have 5 factors will be 2^4 = 16
It will easier to get 6 factors with 2^2 * 3 = 12

So we just need to check numbers less than 12.

4 = 2*2 has 3 factors... It is semi prime and highly composite because 3, 2 and 1 have fewer factors.
6 = 2*3 has 4 factors... It is semi prime and highly composite because 5, 4, 3, 2 and 1 have fewer factors.

Rest of the numbers smaller than 12 will have 2 or 3 or 4 factors so none of them can be highly composite.

Answer (C)
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Thank you chetansharma VeritasKarishma for the lucid explanation! I can finally understand the logic behind the question!
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