## Discussion

Column A | x is an even integer. | Column B |

The number of distinct prime factors of 4x | The number of distinct prime factors of x |

(A) | Column A is larger |

(B) | ... |

(C) | ... |

(D) | ... |

(E) | ... |

(F) | ... |

The solution is

Posted: 10/28/2012 13:04

Isn't Aero an envenenado integer? Couldn't the columns be equal?

Posted: 10/28/2012 13:05

Zero, I mean.

Daniela, 0 is an even integer. Please see the answer explanation. The columns are equal.

Does the justification for eliminating answer "A" come from the "because this is a hard question" reason? Are there any other indicators I've missed that throws out the answer which is essentially the prime #'s in 4X > the prime #'s in X... as far as I can tell answer C only holds water for low values of X.

Hi Bryan,

Yes, Choice A is the 'eye-catcher.' When most people cannot solve a problem, they choose the answer that seems right. But if this were the answer, most students would answer the question correctly and therefore it would not perform as a hard question.

Nova Press

Yes, Choice A is the 'eye-catcher.' When most people cannot solve a problem, they choose the answer that seems right. But if this were the answer, most students would answer the question correctly and therefore it would not perform as a hard question.

Nova Press

I don't mean to nag but I still don't see how choice "A" is eliminated unless it is the mere presence of a variable. I don't recall being told on the test that these are hard questions when they pop up; so other than the variable and not trusting intuition, how do I solve these questions?... forgive my ignorance on this question

Hi Bryan,

Good question: How do we know when a problem is hard?

Exact criteria cannot be given for what makes a problem hard.

It's not based on complexity. There really are no complex problems on the GRE. Even the hard problems are simple, but they are subtle.

As you study GRE problems, you will develop some intuition for the difficulty level of the problems. The exotic nature of this particular problem is a tip off that it is hard.

The purpose of these hard-problem strategies is usually not to solve the problem (though that can occur), rather their purpose is to prevent making an error. If you could not solve the problem, then you should not guess choice (A) because it is too obvious. And if you did solve the problem and got (A) as the answer, then you should check your work for errors and any unwarranted assumptions you may have made.

We solved the problem mathematically by analyzing the equation 4x = (2^2)x and the fact that x is given to be even.

Nova Press

Good question: How do we know when a problem is hard?

Exact criteria cannot be given for what makes a problem hard.

It's not based on complexity. There really are no complex problems on the GRE. Even the hard problems are simple, but they are subtle.

As you study GRE problems, you will develop some intuition for the difficulty level of the problems. The exotic nature of this particular problem is a tip off that it is hard.

The purpose of these hard-problem strategies is usually not to solve the problem (though that can occur), rather their purpose is to prevent making an error. If you could not solve the problem, then you should not guess choice (A) because it is too obvious. And if you did solve the problem and got (A) as the answer, then you should check your work for errors and any unwarranted assumptions you may have made.

We solved the problem mathematically by analyzing the equation 4x = (2^2)x and the fact that x is given to be even.

Nova Press

What does changing 4x to 2^2x tell you? How does that help you solve the problem?

David,

Chose randomly five different even integers, and solve the prime factors for each one of them,

4 , the prime factor is 2

4*4 = 16, the prime factor is 2 so they are equal

------------

6, the prime factors are 2,3

6*4 = 24 the prime factors for 24 = 2 * 2 * 2 * 3 , so only 2, 3

---------------

So based on That the prime factors are equal for what ever number u pick.

Chose randomly five different even integers, and solve the prime factors for each one of them,

4 , the prime factor is 2

4*4 = 16, the prime factor is 2 so they are equal

------------

6, the prime factors are 2,3

6*4 = 24 the prime factors for 24 = 2 * 2 * 2 * 3 , so only 2, 3

---------------

So based on That the prime factors are equal for what ever number u pick.

Its true ,any of even num has one prime factor which is 2 and in example here its given by 4=2*2.so any of way its become a common.

Posted: 02/04/2014 06:07

I do not understand how the answer was derived

Hello Wendy. Two critical information pieces: 1) x is even, and 2) 4 has 2 as its distinct prime factor, in addition to whatever x has as distinct prime factors.

Since x is even, we conclude that 2 is one of its prime factors (e.g., 6 has 2 and 3. 36 has 2, 3, and 9, etc.). Hence having 4 as a multiplier in Column A does not add any additional distinct prime factor to those of x.

The answer then, is C.

Since x is even, we conclude that 2 is one of its prime factors (e.g., 6 has 2 and 3. 36 has 2, 3, and 9, etc.). Hence having 4 as a multiplier in Column A does not add any additional distinct prime factor to those of x.

The answer then, is C.

Posted: 03/15/2017 09:18

If we take 1 as a integer and then 1 is not a prime factor

Hi Arushi,

We cannot choose x to equal 1 because 1 is an odd number, and we are told that 'x is an even integer.'

Nova Press

We cannot choose x to equal 1 because 1 is an odd number, and we are told that 'x is an even integer.'

Nova Press

Posted: 03/15/2017 09:18

So the ans is d