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# If x^2 is divisible by 216, what is the smallest possible value for

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Manager
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If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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23 Jun 2009, 19:35
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If x^2 is divisible by 216, what is the smallest possible value for positive integer x?
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Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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23 Jun 2009, 19:58
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216 = 6X6X6

Now we are asked what is the smallest x such that $$x^2$$ is divisible by 216.
To make 6X6X6 a perfect square, we need one more 6.

Hence 6X6X6X6 = $$36^2$$ is the smallest $$x^2$$ divisible by 216.

What is the OA?
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Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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24 Jun 2009, 08:45
sdrandom1 wrote:

216 = 6X6X6

Now we are asked what is the smallest x such that $$x^2$$ is divisible by 216.
To make 6X6X6 a perfect square, we need one more 6.

Hence 6X6X6X6 = $$36^2$$ is the smallest $$x^2$$ divisible by 216.

What is the OA?

This is correct the OA is 36.

I like your reasoning better than the Guides

Guides explanation
Quote:
The prime box of $$x^2$$contains the prime factors of 216, which are 2, 2, 2, 3, 3, and 3. We know that the prime factors of $$x^2$$ should be the prime factors of x appearing in sets of two, or pairs. Therefore, we should distribute the prime factors of $$x^2$$ into two columns to represent the prime factors of x. We see a complete pair of two 2's in the prime box of $$x^2$$, so x must have a factor of 2. However, there is a third 2 left over. That additional factor of 2 must be from x as well, so we assign it to one of the component x columns. Also, we see a complete pair of two 3's in the prime box of $$x^2$$, so x must have a factor of 3. However, there is a third 3 left over. That additional factor of 3 must be from x as well, so we assign it to one of the component x columns. Thus, x has 2, 3, 2, and 3 in its prime box, so x must be a positive multiple of 36.

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Re: Smallest possible value of x [#permalink]

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26 Oct 2009, 12:42
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216 = 6^3

if we multiply this by 6 we get 6^4 which can be split into a perfect square = 36^2.
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Re: Smallest possible value of x [#permalink]

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26 Oct 2009, 13:41
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Yep, Atish is right, I got 36 too

I almost did the same thing as Atish, except I further split 216 into$$2^3*3^3$$ and for it be a perfect square we need 2*3
so $$2^2*3^2$$ = 36 is the answer
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Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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13 Feb 2011, 11:20
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216 = 2^3 x 3^3

now, (x^2)/216 is an integer

So x must have atles 3 twos and 3 threes

to make it a perfect square, we multiply x^2 with one more 2 and one more 3...

x^2 = 2^3 x 3^3 x 2 x 3 = 2^4 x 3^4

x = sq. rt.(2^4 x 3^4)
x = 2^2 x 3^2
x = 4 x 9
x = 36
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Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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14 Mar 2016, 01:07
I cannot find the options..
Also without looking at the options
as 2,3 are prime factors of x and x^2 is divisible by 216
so x must be at least => 36
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Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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14 Mar 2016, 03:28
Chiragjordan wrote:
I cannot find the options..
Also without looking at the options
as 2,3 are prime factors of x and x^2 is divisible by 216
so x must be at least => 36

No options were provided for this question in the source.
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Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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22 Apr 2016, 10:12
There are Three ways we can solve this question
The Easiest way => track the values i.e. using options start with the smallest value and check if it is divisible by 216 or not . P.S => Choose the least value.
The easier way => x is an integer => x^2 will be perfect square => x=36
Slightly complicated way => x and x^2 have same Prime factors . now x^2 must have 2 and 3 as its prime factors so x= 2*3*p where p is some number
Now x^2 = 36p^2 .
but x^2 is divisible by 216
so p=6 atleast
hence x=36 is the least value of x
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Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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24 Jan 2017, 19:18
1) 216 consists of the following prime factors 2*2*2*3*3*3
2) x should have at least two 2's and two 3's as prime factors in order for its square to be divisible by 216.
3) x=2*2*3*3=36
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If x2 is divisible by 216, what is the smallest possible value for... [#permalink]

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29 Mar 2017, 15:05
If x2 is divisible by 216, what is the smallest possible value for positive integer x?

I came across this problem in one of my study guides. I still don't understand how to do the work even with the explanation provided.

Can someone explain how to tackle this problem?

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If x2 is divisible by 216, what is the smallest possible value for... [#permalink]

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29 Mar 2017, 16:11
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Expert's post
Equinocs wrote:
If x2 is divisible by 216, what is the smallest possible value for positive integer x?

I came across this problem in one of my study guides. I still don't understand how to do the work even with the explanation provided.

Can someone explain how to tackle this problem?

Dear Equinocs,

I'm happy to respond.

First of all, if please use proper math notation. For x-squared, you can write x^2 or $$x^2$$ using the math notation function. Please do not write x2, because this will generate tremendous confusion.

If you happen to know the cubes of all the single digit numbers, that would be easier, because $$6^3 = 216$$. I think all students should have those ten cubes memorized, but let's assume that you didn't know that.

A HUGE help in all kinds of questions about factors and divisible and so forth is the idea of prime factorization. See:
GMAT Math: Factors

Let's start simply by breaking up 216 into factors.
216 = 2*108 = 2*2*54
= 2*2*6*9
= 2*2*2*3*3*3

That's the prime factorization of 216. Whenever you have a particular big number in a question about find the factors or multiples etc, chances are very good that finding the prime factorization of the big number given will be helpful.

Now, another big idea about factors. When we square a number, the factors are doubled: for each single factor, we get a pair of those factors.
$$35 = 5*7$$
$$35^2 = 5*5*7*7$$

$$18 = 2*2*3$$
$$18^2 = 2*2*2*2*3*3$$

Now, let's go back and look at the prime factorization of 216. We can group some of the factors into pairs.
216 = (2*2)*(3*3)*2*3
We have paired and unpaired factors, so to turn this into a perfect square, we would have to multiply by another 2 and another 3, so those factors are paired up.
x^2 = (2*2)*(3*3)*(2*2)*(3*3)
When we take a square-root, each pair of factors becomes a single factor.
x = 2*3*2*3 = 36

Thus, x = 36.

Does this make more sense?
Mike
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Re: If x2 is divisible by 216, what is the smallest possible value for... [#permalink]

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29 Mar 2017, 16:31
Awesome!

Thank you so much! My apologies for the x2 (x^2).
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Re: If x^2 is divisible by 216, what is the smallest possible value for [#permalink]

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29 Mar 2017, 21:30
Equinocs wrote:
If x2 is divisible by 216, what is the smallest possible value for positive integer x?

I came across this problem in one of my study guides. I still don't understand how to do the work even with the explanation provided.

Can someone explain how to tackle this problem?

Merging topics. Please refer to the discussion above.
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Re: If x^2 is divisible by 216, what is the smallest possible value for   [#permalink] 29 Mar 2017, 21:30
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