x=2*3*5*7----59

therefore x is even.

now, since x is even. therefore all the terms in which an even number is added to x will not yield a prime number (because even+even=even which is not prime (except 2) )

now all other odd terms have a common factor in x. for e.g. x+3 has a common factor of 3, x+5 has a common factor of 5......and similarly x+59 has a common factor of 59. thus none of the odd terms yield a prime number.

hence, the given set has a zero prime number in it.

manpreetsingh86 ..... [x+9] doesn't have a common factor ....[9 is not prime] ....i doubt if any product these two primes gives you a multiple of 9.

So that's glitch in the logic here. Anyways i too chose 0 as the answer..but had they given 1, 2, etc...i would have got confused.

Bunuel / VeritasPrepKarishma please help.

x+9 = (2*3*5*7----59 + 9) = 3(2*5*7-----59 +3)
as can be seen 3 is clearly a factor of x+9.

i hope this helps.

Since 2 * 3* 5 *... *59 is even, x is even.
As x is even, so will be x+2, x+4, x+6...... x+58, so these numbers cannot be prime.
Now x+3 is nothing but (2*3*... 59) + 3. Rewriting this as {3 * (2 * 5*.... 59)} +3.
One can observe here that the term in the curly brace is a multiple of 3 and that term plus 3 will definitely be a multiple of 3 .
(Multiple of 3 + 3 is a multiple of 3, Multiple of 59 + 59 is a multiple of 59)
As x+3 is a multiple of 3 it cannot be a prime.
Similarly x+5 can be written as {5 * (2 * 3* 7*.... 59)} +5 which is a multiple of 5.

Like wise we can see that all the terms in the set are multiples of prime numbers.


Ambarish
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Please check the solution

Answer: option A

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