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Divisible by 15 -- Please help

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by Jim@StratusPrep » Tue Feb 14, 2012 6:17 am
Six ally, does x have a factor of 5 and 3?

1). Has a factor of 2 and 5. Insufficient

2). Has a factor of 2,3, and 5. (if x^2 has these factors so does x). Sufficient

B
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by mankey » Thu Feb 16, 2012 10:17 am
IMO: C. Please check.

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by amanne » Thu Feb 16, 2012 9:52 pm
Inorder for X to be divisible by 15, X must have 5 and 3 as factors.

1) X has only 2 and 5 as factors. Insufficient

2) X^2 has only 2,3 and 5 as factors. But X^2 should have at least 2 three's and 2 five's for X to be divisible by 15. Insufficient

Both statements combined are still insufficient.

Answer E.

Can anyone confirm this explanation.
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by [email protected] » Thu Feb 16, 2012 11:23 pm
Is X divisible by 15?

1. When X is divided by 10, the result is an integer.
2. X^2 is a multiple of 30.


Exactly the answer is turning out to be C. Not B.

There are many numbers whose square must be the multiple of 30 but remember their squares are the multiple of 30 X itself...

When both the statements are combined, the answer turns out to be good...
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by sanju09 » Fri Feb 17, 2012 12:27 am
rjanardhanan wrote:Is X divisible by 15?

1. When X is divided by 10, the result is an integer.
2. X^2 is a multiple of 30.
I. This means that X is divisible by 10, hence not necessarily divisible by 15. Insufficient

II. Once X^2 is a multiple of 30, X is a multiple of 30, and hence a multiple of 15 as well. Sufficient

[spoiler]Take B[/spoiler]
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by knight247 » Fri Feb 17, 2012 12:31 am
Hey Amit,
The answer IS B not C.

Is x divisible by 15?
(1)When x is divided by 10, the result is an integer. Ok, so x could be 20,30,40,50,50 ETC all of which meet the requirement of giving an integer result when divided by 10. But x could be 30 which is divisible 15 or it could be 20 which isn't divisible by 15. Conflicting answers so INSUFFICIENT.
(2)x²= some multiple of 30
x²=2*3*5....
Prime factors of perfect squares will obviously come in pairs. So we could rewrite x² as
x²=2²*3²*5²..... And taking square root on both sides
x=2*3*5... Which is divisible by 15. SUFFICIENT.

Hence B
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by sanju09 » Fri Feb 17, 2012 12:32 am
[email protected] wrote:Is X divisible by 15?

1. When X is divided by 10, the result is an integer.
2. X^2 is a multiple of 30.


Exactly the answer is turning out to be C. Not B.

There are many numbers whose square must be the multiple of 30 but remember their squares are the multiple of 30 X itself...

When both the statements are combined, the answer turns out to be good...
Can you please give me a square integer divisible by 30?
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by [email protected] » Fri Feb 17, 2012 1:20 am
Just got stuck a bit here...

Square integer of a multiple of 30 is 900...

And knight247

x^2 = 30, how do u know that it has to be a perfect square... please help me understand this...
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by [email protected] » Fri Feb 17, 2012 1:24 am
X^2 is a multiple of 30.


this can be: x^2 = 30 X 1 = 30
x^2 = 30 X 2 = 60
etc.......


and also x^2 = 30 X 30 this goes , hence insufficient...
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by sanju09 » Fri Feb 17, 2012 1:59 am
[email protected] wrote:X^2 is a multiple of 30.


this can be: x^2 = 30 X 1 = 30
x^2 = 30 X 2 = 60
etc.......


and also x^2 = 30 X 30 this goes , hence insufficient...
If you take X^2 = 30, then X would no more be an integer, and this would contradict the stem which reads the term 'divisible'. Remember, whenever GMAT uses the terms like divisible, a factor of, or a multiple of, they customarily mean integers, preferably non-negative ones.
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by [email protected] » Fri Feb 17, 2012 2:18 am
Ok sanju09! thank you for your support...

I got my mistake...
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by shekhar.kataria » Fri Feb 17, 2012 4:45 am
Hi Sanju
II. Once X^2 is a multiple of 30, X is a multiple of 30, and hence a multiple of 15 as well. Sufficient
How can you prove the above quoted. Just a case which goes opposite to what you said.

6^2 is a multiple of 4, but 6 is not a multiple of 4.
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by Jim@StratusPrep » Fri Feb 17, 2012 9:38 am
shekhar.kataria wrote:Hi Sanju
II. Once X^2 is a multiple of 30, X is a multiple of 30, and hence a multiple of 15 as well. Sufficient
How can you prove the above quoted. Just a case which goes opposite to what you said.

6^2 is a multiple of 4, but 6 is not a multiple of 4.
Please see my original post. This has to do with prime factors. If a number is squared, it has "pairs" for each of the prime factors (see below) - 2 and 3 in the case of 6. To be a multiple of 4 there would have to be 4 factors of 2 in the x^2.

2^2 = 2^2
3^2 = 3^3
4^2 = 2^2 * 2^2
5^2 = 5^2
6^2 = 2^2 * 3^2
7^2 = 7^2
8^2 = 2^2 * 2^2 * 2^2
9^2 = 3^2 * 3^2

Basically, all squared numbers have an even number of prime factors. So if x^2 is a multiple of 30, then x^2 has pairs of prime factors for 2,3 and 5. Thus , so does X.

Hope this clarifies things.
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by ArunangsuSahu » Fri Feb 17, 2012 11:10 am
X^2=30*K

K=1
x^2=30
x=30^1/2 or -30^1/2

Any thought about this?
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