If x and y are distinct prime numbers

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If x and y are distinct prime numbers

by vipulgoyal » Thu Mar 19, 2015 7:47 pm
If x and y are distinct prime numbers, each greater than 2, which of the following must be true?

(I) x+y is divisible by 4
(II)x * y has even number of factors
(III)x+y has an even number of factors

A. I only
B. II only
C. I and III only
D. II and III only
E. I and II only

OE

I. x+y is divisible by 4 --> if x=3 and y=7 then x+y=10, which is not divisible by 4. So this statement is not always true;

II. xy has even number of factors --> only perfect squares have an odd number of factors (check this: a-perfect-square-79108.html?hilit=perfect%20square%20reverse#p791479), as x and y are distinct prime numbers then xy can not be a perfect square and thus can not have an odd number of factors, so xy must have an even number of factors. This statement is always true;

III. x+y has an even number of factors --> now, x+y can be a perfect square, for example if x=3 and y=13 then x+y=16=perfect \ square, so x+y can have an odd number of factors. So this statement is not always true;

Answer: B (II only).


I do have query on OE


Since stem is asking total no to factors NOT differant factore , in this case ans would be D
16 - 2*2*2*2 - total no of factors 16-even

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by GMATGuruNY » Fri Mar 20, 2015 1:58 am
vipulgoyal wrote:I do have query on OE

Since stem is asking total no to factors NOT differant factore , in this case ans would be D
16 - 2*2*2*2 - total no of factors 16-even
Case: x=3, y=13

Here, x+y = 16.
Factors of 16:
1, 2, 4, 8, 16.
Total number of factors = 5.

Since x+y has an ODD number of factors, statement III does NOT have to be true.
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by ceilidh.erickson » Sat Mar 28, 2015 4:29 pm
An integer will have an even number of distinct factors as long as it's not a perfect square.

A prime has exactly two factors: 1 and itself.
A non-prime, non-square will have pairs of factors that multiply to it.
e.g. the factors of 12:
1*12
2*6
3*4

A perfect square, on the other hand, will have an odd number of distinct factors, because we don't want to double-count the square root.
The factors of 36:
1*36
2*18
3*12
4*9
6*6 <-- we only want to count 6 once here, so it's not a pair

So, if you can think of two odd primes (primes greater than 2) that add up to a perfect square, then III will not always be true. Here are some examples:
5+31
3+13
5+59
11+53
29+7
etc
Since stem is asking total no to factors NOT differant factore , in this case ans would be D
16 - 2*2*2*2 - total no of factors 16-even
You seem to be confusing total factors with prime factors here. It is implied in "an even number of factors" that the GMAT means "an even number of DISTINCT factors." Otherwise, we could claim that 16 = 2*2*2*2*1*1*1*1*1... etc. We could count 1 an infinite number of times when factoring any number. Normally, we can't assume that the word "distinct" is implied if not stated, but in the case of "the number of factors," we can - otherwise there wouldn't be a single number.

If a question asked for the number of PRIME factors, then you would indeed break down 16 the way you did.

Did that help?
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by Scott@TargetTestPrep » Tue Jan 09, 2018 9:16 am
vipulgoyal wrote:If x and y are distinct prime numbers, each greater than 2, which of the following must be true?

(I) x+y is divisible by 4
(II)x * y has even number of factors
(III)x+y has an even number of factors

A. I only
B. II only
C. I and III only
D. II and III only
E. I and II only
Since x and y are prime numbers greater than two, they are both odd prime numbers. Let's analyze each answer choice:

I. x+y is divisible by 4

This is not true. If x = 11 and y = 7, then x + y = 18, which is not divisible by 4.

II. x*y has an even number of factors

The only numbers that have an odd number of factors are perfect squares. Since x and y are distinct primes, x*y can't be a perfect square, and thus it must have an even number of factors. This statement is true.

III. x+y has an even number of factors

This is not true. If x = 3 and y = 13, then x + y = 16, which is a perfect square. Recall that in Roman numeral II, we've mentioned that a perfect square has an odd number of factors.

Answer: B

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