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# If x is a positive integer, is sqrt(x) an integer? 1.

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Manager
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If x is a positive integer, is sqrt(x) an integer? 1. [#permalink]  02 Mar 2005, 10:19
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If x is a positive integer, is sqrt(x) an integer?

1. sqrt(4x) is an integer

2. sqrt(3x) is not an integer
Intern
Joined: 02 Mar 2005
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sqrt (4x) = 2 sqrt (x) is an integer

so sqrt (x) is either an integer or a half fraction. If x is integer so sqrt (x) cannot be half fraction.

Therefore sqrt(x) is an integer.
VP
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why the hell did i choose C ??? i have no idea !

1) x must be the multiple of a prime or 1 => suff

2) pick for example 4 it works for stem 2) and for sqrtx; pick 2 it works for stem 2) but definetly not for sqrtx => insuff

no concentration = no success

Last edited by christoph on 02 Mar 2005, 14:24, edited 1 time in total.
VP
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HongHu wrote:
christoph wrote:
1) x must the multiple of a prime or 1 => suff

What do you mean?

x must be a multiple of a prime 2,3,5,7, etc.. or x must be 1 for stem 1) to be true. then sqrtx is a integer ! is it ?

Last edited by christoph on 02 Mar 2005, 14:39, edited 1 time in total.
SVP
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Stem (2)? No, I don't think your reasoning is correct.

toddmartin wrote:
1. sqrt(4x) is an integer

sqrt(4x)=sart(4)*sqrt(x)=2sqrt(x)
If this is an integer, that means either sqrt(x) is an integer, or sqrt is half of an integer (eg, 1/2 or 3/2 etc.). But the stem says that x is an integer, therefore sqrt x cannot be half of an integer. In other words sqrt x must be an integer. (This means x is the square of another integer. It has nothing to do with prime numbers.)
VP
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HongHu wrote:
Stem (2)? No, I don't think your reasoning is correct.

toddmartin wrote:
1. sqrt(4x) is an integer

sqrt(4x)=sart(4)*sqrt(x)=2sqrt(x)
If this is an integer, that means either sqrt(x) is an integer, or sqrt is half of an integer (eg, 1/2 or 3/2 etc.). But the stem says that x is an integer, therefore sqrt x cannot be half of an integer. In other words sqrt x must be an integer. (This means x is the square of another integer. It has nothing to do with prime numbers.)

i mean stem 1), but nevertheless you are right
Manager
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christoph wrote:
HongHu wrote:
christoph wrote:
1) x must the multiple of a prime or 1 => suff

What do you mean?

x must be a multiple of a prime 2,3,5,7, etc.. or x must be 1 for stem 1) to be true. then sqrtx is a integer ! is it ?

The OA is A. It has nothing to do with prime numbers. In fact, prime numbers wouldn't work with statement 1.

I looked at it like this: sqrt(4x)=sqrt(4)*sqrt(x). St1 says that this is an integer. Since sqrt(4)=2, then 2 * sqrt(x)= a postitive integer. In order for it to work sqrt(x) must be an integer.

I'm sure you all have the OG so you can look up the offical explanation.
VP
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toddmartin wrote:
christoph wrote:
HongHu wrote:
christoph wrote:
1) x must the multiple of a prime or 1 => suff

What do you mean?

x must be a multiple of a prime 2,3,5,7, etc.. or x must be 1 for stem 1) to be true. then sqrtx is a integer ! is it ?

The OA is A. It has nothing to do with prime numbers. In fact, prime numbers wouldn't work with statement 1.

I looked at it like this: sqrt(4x)=sqrt(4)*sqrt(x). St1 says that this is an integer. Since sqrt(4)=2, then 2 * sqrt(x)= a postitive integer. In order for it to work sqrt(x) must be an integer.

I'm sure you all have the OG so you can look up the offical explanation.

why wouldn`t prime numbers work with stem 1 ? to be precise i said mulitples of primes. IMO 9 or 4 or 25 would verify stem 1 as well as sqrtx. what i meant instead of multiples of primes is multiples of the products of different primes. but in fact i was wrong.
Director
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(A)

if the product of 4 and any number yields a number that has a sqrt, then that number must have a sqrt
Senior Manager
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all squares are multiples of primes but not all multiples of primes are squares
for example
4 is multiple of 2
but 10 is multiple of 2 and 5 (primes) but it isn't a square (of an integer)
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