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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Kindly press Kudos if the explanation is clear.
Thank you
Ambarish
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71% (01:17) correct 
