If x is an integer,

[bgussin]

I searched and couldnt find this one on the boards already. The question goes:

If x is an integer, is (x^2 + 1)(x+5) and even number?

1) x is an odd number
2) Each prime factor of x^2 is greater than 7

I know that 1 is sufficient, because 2 evens mult together gives you an even number. I am not sure how to figure out if 2 is sufficient, which it is by the way. Any help would be greatly appreciated. Thanks!

[ctrajaram]

If we prove x^2+1 is even or x+5 is even or both then we know the product is even

You got stm I

Coming to stmt II

This is indirectly stating that x is odd. If x^2 prime factors are greater than 7 then 2 is not a prime factor of x^2 and therefore x so we know x is odd

SUFF

[RonPurewal]

I searched and couldnt find this one on the boards already. The question goes:

If x is an integer, is (x^2 + 1)(x+5) and even number?

1) x is an odd number
2) Each prime factor of x^2 is greater than 7

I know that 1 is sufficient, because 2 evens mult together gives you an even number. I am not sure how to figure out if 2 is sufficient, which it is by the way. Any help would be greatly appreciated. Thanks!

statement 2 is just being obnoxious; they're testing you to see whether you can decode this statement properly, and get down to the essence of what it's trying to tell you.

first of all, an important takeaway that seems to recur a lot:
POWERS of a number have EXACTLY THE SAME PRIME FACTORS as does the ORIGINAL NUMBER.
reason:
think about how you create powers: you just take a number, and multiply together multiple copies of the same number.
by so doing, you're just repeating the same prime factors, over and over and over again.

so, in this context, "prime factors of x^2" is the same as just "prime factors of x".

therefore,
(2) each prime factor of x is greater than 7

at this point, you should be thinking about even and odd, even though even/odd is not specifically addressed by statement 2.
you should be thinking about even/odd anyway, even though they are not mentioned in the statement, because the REST OF THE PROBLEM is clearly related to even/odd.

since ALL primes greater than 2 (and thus, a fortiori, all primes greater than 7) are odd, we have
each prime factor of x is odd
and therefore
x is odd.

this statement is therefore sufficient for the same reasons as is statement (1).