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

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Director
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Re: If x is an integer, is (x^2+1)(x+5) an even number?  [#permalink]

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29 Jan 2017, 17:33
If x is an odd number, x^2 must be odd => (x^2+1) must be even.
By the same logic, (x+5) is even. As addition of two odd numbers results in an even number.
Multiplication of two even numbers results in an even number. Thus (x^2+1)(x+5) is an even number.

Satetment (1) is SUFFICIENT.

And for (2), each prime factor of x^2 is greater than 7 implies that 2 is not a prime factor of x^2 and in turn 2 is not a prime factor of x. (As all the prime factors of x^2 are also prime factors of x)

That means x is composed of 11, 13, 17, 19, 23 etc (Prime numbers greater than 7).
Therefore x is an odd number => Equivalent to statement (1).

Statement (2) is SUFFICIENT.

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If x is an integer, is (x^2+1)(x+5) an even number?  [#permalink]

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13 Jul 2017, 00:11
Rephrase the question.

The original question is: Is $$(x^2+1)(x+5)$$ even?

This is the case when EITHER $$(x^2+1)$$ AND/OR $$(x+5)$$ are even.

$$(x^2+1)$$ is even when "x" is odd. $$(x+5)$$ is also even when "x" is odd.

Therefore, the new question is: Is "x" odd?

A.) X is odd. Well.... that's a layup. SUFFICIENT.

B.) Each prime factor of $$x^2$$ is greater than 7

That's a fancy way of saying that "x" is odd.

Any number that has "2" as a factor is even, and any number that doesn't is odd. Every prime number greater than "2" is odd. If the statement is telling us that $$x^2$$ is built up of a bunch of prime numbers greater than 7, then all of those prime numbers are odd.

And the prime factorization of "x" will just be half the number of the same odd primes that made up $$x^2$$.

For any number to be even, "2" has to be one of its prime factors. This statement is telling you that isn't the case.

So therefore, we know that "x" is odd.

SUFFICIENT.

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Re: If x is an integer, is (x^2+1)(x+5) an even number?  [#permalink]

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17 Jul 2017, 14:01
mybudgie wrote:
If x is an integer, is (x^2 + 1)(x + 5) an even number?

(1) x is an odd number

(2) Each prime factor of x^2 is greater than 7

Before looking at each statement, we could first try to understand the expression- an odd squared plus an odd is even; an odd plus odd is even- an even times even is even; however, an even squared plus an odd is odd AND an odd plus an even is odd. The question then becomes- is x odd?

Statement 1
Suff- clearly tells us that x is an odd

Statement 2
This statement just implies that 2 is not a factor of x^2- so the number could be 11^2 or 13^2- in either case this statement requires an odd number in order to be true.

D
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Re: If x is an integer, is (x^2+1)(x+5) an even number?  [#permalink]

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19 Jan 2018, 11:28
In statement 2. Can X not be 0?
each prime number of X^2 is greater than 7. Understanding that 2 is the only prime number so if the prime factor of x^2 is above 7 then the only numbers we are left with are odd numbers. Since 0 a multiple of every number. Can x be 0?

Maybe it's too dumb of a question but hey we are all here to learn. Thanks
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Re: If x is an integer, is (x^2+1)(x+5) an even number?  [#permalink]

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19 Jan 2018, 11:34
Rocket7 wrote:
In statement 2. Can X not be 0?
each prime number of X^2 is greater than 7. Understanding that 2 is the only prime number so if the prime factor of x^2 is above 7 then the only numbers we are left with are odd numbers. Since 0 a multiple of every number. Can x be 0?

Maybe it's too dumb of a question but hey we are all here to learn. Thanks

x cannot be 0 because 0 is divisible be primes which are less than 7 (2, 3, and 5).
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Re: If x is an integer, is (x^2+1)(x+5) an even number? &nbs [#permalink] 19 Jan 2018, 11:34

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