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↧↧↧ Detailed Video Solution to the Problem ↧↧↧


x# is defined for every positive even integer x as the product of all even integers from 2 to x

Lets take an example values of x
Lets say x = 4
4# = 2*4 = 8

What is the smallest possible prime factor of (x#+7)?

Now, x# will always be an even number as it is product of even numbers
=> x# + 7 = Even + Odd = Odd
=> 2 cannot be the factor
=> Eliminate A

Lets take values of x and test. Let x be 4 itself
=> 4# + 7 = 8 + 7 = 15
=> 3 is a factor

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

­
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Is this question valid for all even numbers???!!!
What about when x = 8?

8# = 2 * 4 * 6 * 8 = 384
8# + 7 = 391
And 391's prime factorization is 17 * 23.

In that case, none of the options are suitable!!! And unfortunately this is the example I took.
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Is this question valid for all even numbers???!!!
What about when x = 8?

8# = 2 * 4 * 6 * 8 = 384
8# + 7 = 391
And 391's prime factorization is 17 * 23.

In that case, none of the options are suitable!!! And unfortunately this is the example I took.
Check the question again: “What is the smallest possible prime factor of (x# + 7)?”

It asks for the smallest prime factor that is possible, not one that must occur for every even x. When x = 2, x# + 7 = 2 + 7 = 9, whose smallest prime factor is 3.
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Ah, I get it. This is the second question in my today's practice that has tripped me up using this kind of round-about words.... Phew...😓
Bunuel

Check the question again: “What is the smallest possible prime factor of (x# + 7)?”

It asks for the smallest prime factor that is possible, not one that must occur for every even x. When x = 2, x# + 7 = 2 + 7 = 9, whose smallest prime factor is 3.
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