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Is positive integer x divisible by 24?

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Is positive integer x divisible by 24?  [#permalink]

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New post 23 Jan 2014, 12:50
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Is positive integer x divisible by 24?

(1) √x is divisible by 4.
(2) x^2 is not divisible by 9.
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Re: Is positive integer x divisible by 24?  [#permalink]

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New post 23 Jan 2014, 12:52
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Prime factors of 24: 2^3 * 3^1. In order for a number (x) to be divisible by 24, it needs to have all these prime factors at least.
(It may or may not have additional factors).

Statement 1: Square on both sides (imagine this is a equation). This yields that x is divisible by 16.
This means: x = k*16 for some constant k. Now, 16 = 2^4 which is more than 2^3. But no 3. In case k includes at least a 3,
x is divisible by 24. In case k does not include a 3, x is not divisible by 24. INSUFFICIENT

Statement 2: x^2 is not divisible by 9. Again imagine an equation, but take now the square root on both sides.
This means that x is not divisible by 3. This also means that x does not have a single 3 as one of its prime factors.
But in order to be divisible by 24, we know it needs at least 3^1 (i.e. at least one 3). Thus, we know that x cannot be
divided by 24 as it lacks 3 as its prime factor. SUFFICIENT

Answer is B.
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Re: Is positive integer x divisible by 24?  [#permalink]

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New post 27 Jan 2014, 00:47
BabySmurf wrote:
Is positive integer x divisible by 24?

(1) √x is divisible by 4. (this means to read square root of x is divisible by 4)
(2) x^2 is not divisible by 9.


All DS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=354
All PS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=185


Hope this helps.
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Re: Is positive integer x divisible by 24?  [#permalink]

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New post 03 Oct 2014, 10:40
Can someone please explain why statement 1 is not suffificent. If x=16, not divisible by 24. x=64, not divisible by 24. x=144, not divisible by 24 and so on...
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Re: Is positive integer x divisible by 24?  [#permalink]

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New post 03 Oct 2014, 10:46
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aadikamagic wrote:
Is positive integer x divisible by 24?

(1) √x is divisible by 4.
(2) x^2 is not divisible by 9.

Can someone please explain why statement 1 is not suffificent. If x=16, not divisible by 24. x=64, not divisible by 24. x=144, not divisible by 24 and so on...


144 is divisible by 24: 144/24 = 6.
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Re: Is positive integer x divisible by 24?  [#permalink]

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New post 08 Jul 2017, 12:37
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BabySmurf wrote:
Is positive integer x divisible by 24?

(1) √x is divisible by 4.
(2) x^2 is not divisible by 9.


In order to check the divisibility by x first lets check the factors of \(24\)

\(24 = 3 * 8 = 3 * 2 * 2 * 2 = 3^1 * 2^3\)

Hence, For any number of be divisible by 24 we will need one 3 and three 2's.

Lets check the options.

(1) \(\sqrt{x}\) is divisible by \(4\)

\(= \frac{\sqrt{x}}4\)

Squaring Numerator and Denominator

\(= \frac{x}{16} = \frac{x}{2*2*2*2} = \frac{x}{2^4}\)

This means x = k * x is divisible by 2^3

If the value of k has 3 then x is divisible by 24

However, if the value of k does not have 3 then x is not divisible by 24

Hence, (1) ===== is NOT SUFFICIENT


(2) \(x^2\) is not divisible by 9

As \(x^2\) is not divisible by 9, x will not be divisible by 3.

In order for a number to be divisible by 24, we need at least one prime number - 3, and as x is not divisible by 3 we can say that x is not divisible by 24.

As we are able to answer the question - x divisible by 24? (2) is SUFFICIENT

Hence, Answer is B
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Re: Is positive integer x divisible by 24?  [#permalink]

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New post 08 Jul 2017, 14:54
Accidentally hit C but the answer is B

Prime Factorization of 24 = 2^3 * 3

A) Tells us that root X is divisible by 4, meaning that X will have at least 4 2s (4*4), making it divisible by 8. There is no mention of 3, so this is not sufficient.

B) Tells us that x^2 is not divisible by 9. If 3 was a factor of X then x^2 would be divisible by 9 (3*3). Thus, X cannot possibly be divisible by 24 since 3 is not a factor of X

Hope this helps
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Re: Is positive integer x divisible by 24?  [#permalink]

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New post 28 Aug 2018, 06:02
BabySmurf wrote:
Is positive integer x divisible by 24?

(1) √x is divisible by 4.
(2) x^2 is not divisible by 9.

\(x \geqslant 1\,\,\operatorname{int}\)

\(\frac{x}{{{2^3} \cdot 3}}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int}\)

\(\left( 1 \right)\,\,\,\,\frac{{\sqrt x }}{{{2^2}}}\,\,\, = \,\,\,\operatorname{int} \,\,\,\,\,\,\left\{ \begin{gathered}
\,Take\,\,x = {2^4}\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\,Take\,\,x = {\left( {{2^3} \cdot 3} \right)^2}\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,{\left( {\frac{x}{3}} \right)^2} = \frac{{{x^2}}}{{{3^2}}}\,\,\, \ne \,\,\,\operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{x}{3}\,\,\, \ne \,\,\,\operatorname{int} \,\,\,\,\,\,\,\mathop \Rightarrow \limits^{FOCUS\,!} \,\,\,\,\,\frac{x}{3}\,\,\, \ne \,\,\,{2^3} \cdot \operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{x}{{{2^3} \cdot 3}}\,\,\, \ne \,\,\,\operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle\)

The above follows the notations and rationale taught in the GMATH method.
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Re: Is positive integer x divisible by 24? &nbs [#permalink] 28 Aug 2018, 06:02
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