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If the integer n is greater than 1, is n = 2? 1. n has

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If the integer n is greater than 1, is n = 2? 1. n has [#permalink] New post 16 Aug 2005, 12:27
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If the integer n is greater than 1, is n = 2?

1. n has exactly 2 positive factors.
2. The difference of any 2 distinct positive factors of n is odd.

Posting another one here....

If x and y are +ve integers such that x = 8y + 12, what is the greatest common divisor of x and y?

1. x = 12u, where u is an integer.
2. y = 12z, where z is an integer.
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Re: PS - factors [#permalink] New post 16 Aug 2005, 12:32
Priti wrote:
If the integer n is greater than 1, is n = 2?

1. n has exactly 2 positive factors.
2. The difference of any 2 distinct positive factors of n is odd.


IMO B

COndition 1: could be any prime number

Condition 2: Cannot be 2...because two positive factors of 2 are 2 and 1 and difference is 2 -1 = 1.

Question 2, IMO B again

COndition 1, HCF could be 4 or 12
COnditon 2, HCF could only be 12...
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Last edited by ranga41 on 16 Aug 2005, 18:14, edited 1 time in total.
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 [#permalink] New post 16 Aug 2005, 12:46
I will go with C on the first one.
Statement 1: Can be any prime number
Statement 2: n can be 10 or 2

Both put together n is a prime number, to satisfy statement 2 n has to be 2
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 [#permalink] New post 16 Aug 2005, 13:04
Since: "The difference of any 2 distinct positive factors of n is odd"
10 cannot match here: 5-1=even
I would go with B - only 2 fits.
Correct me if I am wrong...
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 [#permalink] New post 16 Aug 2005, 13:40
Excellent questions, Priti (not only those two - in general)
=================

1. I agree with ranga41 on a tricky B

Statement (1) tells us that n is a prime number, so could be 2,3,5, so on. Not sufficient.

Statement (2)... I am evaluating two sets: all the prime numbers and all the non-prime integers >2

Among the prime numbers only 2 fulfills the stated condidtion, because
3-1=2 5-1=4 and so on...

All the non-prime integers>2 can be broken down to primes, so the simplest of them will have at least 4 factors: 1, x, y and xy, where x and y are primes.

If x and y are even, than xy - 1=odd, but y-x is even => not ok
If x is even and y is odd, than xy-1=odd, but y-1 is even => not ok
If x is odd and y is even, same as above => not ok
If x and y are odd, than all the differences are even => not ok

Since we have run out of numbers to test, 2 seems to be the only number which satisfies statement (2), so the answer should be B.

===============

I will go with a tentative B on the second question. Here is my reasoning:

Statement (1) gives us information about x, but nothing about y => not sufficient.

Statement (2) tells us that 12 is a factor of y. If we substitute into the stem, we get

x=8*12z +12=12(8z+1)

So we need to find the greatest common divisor of x=12(8z+1) and y=12z.

12 is obviously a commont divisor, but what about (8z+1) and z? Тheir only common divisors are 1 and -1, because (8z+1)/z=8+1/z.

To sum it up, I think the second statement gives us enough information and I pick answer choice B
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 [#permalink] New post 16 Aug 2005, 17:07
well..for (1)
I think it is C, cause

in (II) difference of any postive factors is odd...well i read it as say factors of 6 are 1, 2, 3, 6, in this case 6-3=3 which is odd, I am picking any 2 distinct factors...

combining the two statements we know that there will be only 2 positive factors and that there difference is odd...clearly it is 2 then...

yaron wrote:
Since: "The difference of any 2 distinct positive factors of n is odd"
10 cannot match here: 5-1=even
I would go with B - only 2 fits.
Correct me if I am wrong...
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Re: PS - factors [#permalink] New post 16 Aug 2005, 17:53
Priti wrote:
If the integer n is greater than 1, is n = 2?

1. n has exactly 2 positive factors.
2. The difference of any 2 distinct positive factors of n is odd.



1. n has exactly 2 factors, that would be 1 and n itself. So I agree that any primary number would fit the bill. Insuficient.

2. Since 1 and n would be two of the factors. Therefore n-1 is odd. In other words, n must be even. But if n is an even number that is greater than 2, then 2 must be its factor too. And n-2 would NOT be odd. In other words, n has to be an even number that is not greater than 2. That is, n=2. Sufficient.

I'm choosing B.

Quote:
If x and y are +ve integers such that x = 8y + 12, what is the greatest common divisor of x and y?

1. x = 12u, where u is an integer.
2. y = 12z, where z is an integer.


1. x=12u
x=8y+12=12(2/3y+1)
=> 2/3y is an integer.
In other words y is divisible by 3. Since x is also divisible by 3, 3 is one of the common divisors of x and y. However, is it the greatest? We can't tell. For example, if y is divisible by 2, then 6 could be one of the common divisors. And we still don't know if it is the greatest.
Insufficient

2. y=12z
x=8y+12=12*(8z+1)
Since 1 would be the greatest common divisor of z and (8z+1), we know for sure that the greatest common divisor of x and y is 12.
Sufficient

B
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Re: PS - factors   [#permalink] 16 Aug 2005, 17:53
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