218. 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.
Solution:
[spoiler](A) This tells us that x is prime (it's actually the definition of a prime number)
So, x could equal 2, but it could also equal 3, 5, 7 etc --> INSUFFICIENT
(B) Factor sum is odd. So, x could equal 2 (1+2=3) or x could equal 4 (1+2+4=7)--> INSUFFICIENT
(A&B): (A) tells us that x is a prime number. Now most prime numbers (except 2) are odd, which means the two factors will be odd (1 and the prime number itself), so the sum of the factors will always be even (e.g., x=7 --> factors are 1 and 7 --> sum=8)
However, the two factors of 2 (1 and 2) add to be an odd number (3). So, x must equal 2, which means(A)&(B) are sufficient.
Answer = C[/spoiler]
pl explain ..why/how B is not sufficient.
B..says The sum of any two distinct positive factors of x is odd.
while explanation says- b is insufficient because B) Factor sum is odd. So, x could equal 2 (1+2=3) or x could equal 4 (1+2+4=7)--> INSUFFICIENT
is it satisfy the condition : sum of any two distinct positive factors of x is odd...??
1+2=3 odd.... ok
1+4=5 odd ... ok.
2+4=6...even no...
why it is 1+2+4=7..odd... why summimg all three...??
pl explain.
thanks.
i doubt....pl explain. 300+ gmat
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- Rahul@gurome
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218. 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
Obviously A) is the definition of a prime number and so it can take any value 2, 3, 5, 7....
Next consider B) alone.
Now x is either odd or even.
If x is odd, 1 being one of its factors, x+1 will be even. This is against the condition in B).
So x cannot be odd. It has to be even.
Let x = 2n.
So x can have factors 1, 2, n.
Here n can take values 1, 2, 3..
If n is 1, x is 2 and the distinct factors of x are 1, 2.
Since 1+2 = 3 is odd, condition B) is satisfied for x = 2.
Let us see if we can get any other value of x.
Now other values of n can be 2, 3, 4...
If n is odd it will add up with 1 to give an even value and if it is even it will add up with 2 to give an even value.
So n cannot take any value > 1.
It can only be 1 and hence x can only be 2n = 2.
So B) alone is sufficient to answer the question.
A) x is evenly divisible by exactly two positive integers.
B) The sum of any two distinct positive factors of x is odd
Obviously A) is the definition of a prime number and so it can take any value 2, 3, 5, 7....
Next consider B) alone.
Now x is either odd or even.
If x is odd, 1 being one of its factors, x+1 will be even. This is against the condition in B).
So x cannot be odd. It has to be even.
Let x = 2n.
So x can have factors 1, 2, n.
Here n can take values 1, 2, 3..
If n is 1, x is 2 and the distinct factors of x are 1, 2.
Since 1+2 = 3 is odd, condition B) is satisfied for x = 2.
Let us see if we can get any other value of x.
Now other values of n can be 2, 3, 4...
If n is odd it will add up with 1 to give an even value and if it is even it will add up with 2 to give an even value.
So n cannot take any value > 1.
It can only be 1 and hence x can only be 2n = 2.
So B) alone is sufficient to answer the question.
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- Tani
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The only way to get an odd number by summing two others is to sum an odd and an even. To have the sum of "any two" factors be odd, you would always have to be adding an odd and an even. If you are dealing with a non-prime, you will always have 1 as a factor.
If the non-prime is even, you will also have 2 as a factor. No matter what the other factors are (and you must have other factors since the number isn't prime) you will get either odd or even depending on whether you add 1 or 2.
If the non-prime is odd, you must have at least two factors other than 1 and the other factors must also be odd. If you add 1 to any of them you will get an odd number. iI you add two of the odd factors you will get an even number. Therefore, you can get an odd or an even.
The only case in which you will always get an odd number is the one in which you only have two factors, one odd and one even - that means 2.
If the non-prime is even, you will also have 2 as a factor. No matter what the other factors are (and you must have other factors since the number isn't prime) you will get either odd or even depending on whether you add 1 or 2.
If the non-prime is odd, you must have at least two factors other than 1 and the other factors must also be odd. If you add 1 to any of them you will get an odd number. iI you add two of the odd factors you will get an even number. Therefore, you can get an odd or an even.
The only case in which you will always get an odd number is the one in which you only have two factors, one odd and one even - that means 2.
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Here are facts that I understood -
1. To get an odd number out of of the sum of 2 distinct factors is that 1 has to be odd and the other has to be even.
But why is 1+4 (x=4) not possible??
Here again 1 and 4 are two distinct factors of the integer 4 and they sum up to 5 (odd as mentioned by B)
Some one please explain!!
1. To get an odd number out of of the sum of 2 distinct factors is that 1 has to be odd and the other has to be even.
But why is 1+4 (x=4) not possible??
Here again 1 and 4 are two distinct factors of the integer 4 and they sum up to 5 (odd as mentioned by B)
Some one please explain!!
- goyalsau
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The sum of any two distinct positive factors of x is odd.Deepthi Subbu wrote:Here are facts that I understood -
1. To get an odd number out of of the sum of 2 distinct factors is that 1 has to be odd and the other has to be even.
But why is 1+4 (x=4) not possible??
Here again 1 and 4 are two distinct factors of the integer 4 and they sum up to 5 (odd as mentioned by B)
Some one please explain!!
Take the example of 4
Factors of 4 will be 1 , 2 , 4
in all we can make 3 pairs,
1 + 2 = 3 ( odd ) 2 + 4 = 6 ( even ) 1 + 4 = 5 ( odd )
With the help of this reasoning we can eliminate all positive even multiples of 2 , Because in every we will have at least two factors which will be even one is 2 and other is the number itself.
Now Odd Numbers - 1 is factor of every +ve integer so 1 + number it self = will always be equal to even .
All odd numbers can be discarded
Now we have only 1 and 2 left
1 has only 1 +ve factor that is 1 , { we need at least 2 factors }
2 has two factors 1 and 2 which will is odd,
so 2 is the answer...
Saurabh Goyal
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