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300 Questions: Prime

Expert replies
by ashish2104 » Mon Aug 16, 2010 1:13 am
If the integer x is greater than 1, does x = 2?
A) x is evenly divisible by exactly two positive integers.
B) The sum of any two distinct positive factors of x is odd.

Ans given is C, but i think B is sufficient. Only number 2 has its sum of 2 distinct factors odd.
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Source: — Data Sufficiency |

by sanju09 » Mon Aug 16, 2010 1:20 am
ashish2104 wrote:If the integer x is greater than 1, does x = 2?
A) x is evenly divisible by exactly two positive integers.
B) The sum of any two distinct positive factors of x is odd.

Ans given is C, but i think B is sufficient. Only number 2 has its sum of 2 distinct factors odd.
B is the right answer to me too. Never worry, if an integer greater than 1 has sum of any two distinct positive factors of it as odd, it cannot have more than two factors plus it got to be even too. This x is just [spoiler]2.

B
[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by alcatch » Mon Aug 16, 2010 3:01 am
No, you need to read the second statement on its own - if the number is a multiple of two and another odd factor, eg 3, then 1+6 and 2+3 are odd. So it could be 2 or 6. Also, 5 and 2 multiply to make 10; add to make 7; 10 and 1 make 11... get the picture?

C is good.
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by Gurpinder » Mon Aug 16, 2010 10:27 am
ashish2104 wrote:If the integer x is greater than 1, does x = 2?
A) x is evenly divisible by exactly two positive integers.
B) The sum of any two distinct positive factors of x is odd.
Actually, the answer (C) makes sense.

Stmt 1:
This is insufficient because this is true for any prime number.

Stmt 2:
The sum of divisors of ANY odd number will be even. Therefore, if the sum is Odd, we can only consider even numbers.

So lets say that X is 2. 2 is divisible by 2,1 which = 3. This works for Stmt 2.
But lets say that X is 4. 4 is divisible by 4,1 which = 5. This is also an ODD number.

Therefore Stmt 2 is insufficient by itself.

Together.
X is a prime number. And 2 is the only prime number for which the sum of its divisors = odd. The rest are all even.

I hope this helps!
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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by vijaynaik » Mon Aug 16, 2010 11:16 pm
@Gurpinder, in your example 4 not only has 1,4 but also has 2.

2nd statement says sum of any 2 distinct factors: so for 4 4+2 is not odd similarly for 5,6..

So i think B should be correct .
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by uwhusky » Mon Aug 16, 2010 11:27 pm
B can't be correct.

If the integer x is greater than 1, does x = 2?
2) The sum of any two distinct positive factors of x is odd.

x could be anything.

Let's use 10, factors of 10 are 1, 2, 5, 10; 1 + 2 = 3.

Let's try 12, factors of 12 are 1, 2, 3, 4, 6, 12; 1 + 6 = 7.

So 2) says that x could be either 10 or 12, and neither of them is 2.
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by sanju09 » Tue Aug 17, 2010 5:18 am
uwhusky wrote:B can't be correct.

If the integer x is greater than 1, does x = 2?
2) The sum of any two distinct positive factors of x is odd.

x could be anything.

Let's use 10, factors of 10 are 1, 2, 5, 10; 1 + 2 = 3.

Let's try 12, factors of 12 are 1, 2, 3, 4, 6, 12; 1 + 6 = 7.

So 2) says that x could be either 10 or 12, and neither of them is 2.

Please read Statement 2 without falling in to the trap there. It reads sum of any two distinct positive factors of x is odd; any two means any two. If you try any number other than 2, you will find the core condition as violated.

First try x for some prime NOT 2, it got to be an odd now, the only and other factor, 1, is already odd. The sum cannot be odd. Hence the author is definitely not calling x as an odd prime.

Next try any composite you please and see how the core condition is violated.

Try 10, factors being 1, 2, 5, 10; since any two means any two, why shouldn't we try 2 + 10 = EVEN to prove the core condition as violated. Or try 12, factors being 1, 2, 3, 4, 6, 12; since any two means any two, why shouldn't we try 2 + 4 = EVEN to prove the core condition as violated again. You can keep trying if you cannot admit the fact that to any composite, the factors are either all ODD or at least two EVEN.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by gmatmachoman » Tue Aug 17, 2010 5:29 am
@Sanju,

I am with u...B is very much sufficient.. well explained Sanju....
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by uwhusky » Tue Aug 17, 2010 8:23 am
sanju09 wrote:
uwhusky wrote:B can't be correct.

If the integer x is greater than 1, does x = 2?
2) The sum of any two distinct positive factors of x is odd.

x could be anything.

Let's use 10, factors of 10 are 1, 2, 5, 10; 1 + 2 = 3.

Let's try 12, factors of 12 are 1, 2, 3, 4, 6, 12; 1 + 6 = 7.

So 2) says that x could be either 10 or 12, and neither of them is 2.

Please read Statement 2 without falling in to the trap there. It reads sum of any two distinct positive factors of x is odd; any two means any two. If you try any number other than 2, you will find the core condition as violated.

First try x for some prime NOT 2, it got to be an odd now, the only and other factor, 1, is already odd. The sum cannot be odd. Hence the author is definitely not calling x as an odd prime.

Next try any composite you please and see how the core condition is violated.

Try 10, factors being 1, 2, 5, 10; since any two means any two, why shouldn't we try 2 + 10 = EVEN to prove the core condition as violated. Or try 12, factors being 1, 2, 3, 4, 6, 12; since any two means any two, why shouldn't we try 2 + 4 = EVEN to prove the core condition as violated again. You can keep trying if you cannot admit the fact that to any composite, the factors are either all ODD or at least two EVEN.
You're right, the key word here is "any".
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