If x is an integer , is (x^2+1)(x+5) even number?
1. x is odd number
2. each prime factor of x^2>7
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aleph777
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Great question, Gana!
First, let's rephrase the question. If you think strictly in terms of odds and evens, what do you need when multiplying numbers in order to create an even? You just need one even! So initially you need to know if either (x^2 +1) OR (x+5) is even. But we can simplify this one step further. In order for x^2 +1 to be even, x needs to be odd (because when adding, only an odd + odd or even + even will yield an even). And the same goes for x+5: in this case, x needs to be odd.
So you can very quickly simplify the question stem to this: Is X odd?
Now we can move on to the statements.
Statement 1: X is odd. Right there, we confirm the simplified question. SUFFICIENT.
Statement 2: Each prime factor of x^2 > 7. This one's a bit more difficult, but remember that the only even prime number is 2. Every other prime is odd. So now we know that x^2 is made up of only odd factors. Which means x is made up of only odd factors. And since you can only yield an even product by multiplying by AT LEAST one even number, this means x is odd, since it has only prime factors. Therefore, SUFFICIENT.
Thus, D
First, let's rephrase the question. If you think strictly in terms of odds and evens, what do you need when multiplying numbers in order to create an even? You just need one even! So initially you need to know if either (x^2 +1) OR (x+5) is even. But we can simplify this one step further. In order for x^2 +1 to be even, x needs to be odd (because when adding, only an odd + odd or even + even will yield an even). And the same goes for x+5: in this case, x needs to be odd.
So you can very quickly simplify the question stem to this: Is X odd?
Now we can move on to the statements.
Statement 1: X is odd. Right there, we confirm the simplified question. SUFFICIENT.
Statement 2: Each prime factor of x^2 > 7. This one's a bit more difficult, but remember that the only even prime number is 2. Every other prime is odd. So now we know that x^2 is made up of only odd factors. Which means x is made up of only odd factors. And since you can only yield an even product by multiplying by AT LEAST one even number, this means x is odd, since it has only prime factors. Therefore, SUFFICIENT.
Thus, D
