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S is composed of 2 prime integers.... q^3 cannot give a prime integer (ex 11^3, 13^3 are not prime integers), so it must be equal to 1 to not interfer and to make it possible for S to be a multiple of 2 primes integers, provided by p here.

This is abad question though. When combining the 2 statements we need to assume both are true. So by stmt 1 we are told that S is a product of 2 prime ints > 10. Using stmt2, q^3 cannot exist at all

This is abad question though. When combining the 2 statements we need to assume both are true. So by stmt 1 we are told that S is a product of 2 prime ints > 10. Using stmt2, q^3 cannot exist at all

ST. 1 says S is the product of two prime numbers which are both >10
ST. 2 gives us S= P*Q^3

which means P is a prime>10
and Q^3 is a prime>10

hence Q^3 needs to be a perfect cube + a prime> 10
perfect cubes >10 are 27, 64, 125, none of which are primes
there is no perfect cube which is also a prime> 10
how can we take Q to be equal to 1, that would make Q^3= 1 and 1<10

S is composed of 2 prime integers.... q^3 cannot give a prime integer (ex 11^3, 13^3 are not prime integers), so it must be equal to 1 to not interfer and to make it possible for S to be a multiple of 2 primes integers, provided by p here.

Ah ok now I see. I thought S1 was sayin that S is the product of 2 primes that are both greater than 10.

S is composed of 2 prime integers.... q^3 cannot give a prime integer (ex 11^3, 13^3 are not prime integers), so it must be equal to 1 to not interfer and to make it possible for S to be a multiple of 2 primes integers, provided by p here.

Ah ok now I see. I thought S1 was sayin that S is the product of 2 primes that are both greater than 10.

Was like how can this be C?

exactly, I am not sure why everyone is agreeing that Q is 1 when ST. 1& 2 implies clearly that Q^3 must be a prime> 10.
1 is not a prime and it is not greater than 10.