Is positive integer x divisible by 24?

Question

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

BabySmurf · Accepted Answer

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.

Bunuel · Answer

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

aadikamagic · Answer

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...

Bunuel · Answer

144 is divisible by 24: 144/24 = 6.

ydmuley · Answer

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

gmathopeful19 · Answer

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

fskilnik · Answer

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

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

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

\left( 2 ight)\,\,\,{\left( {\frac{x}{3}} ight)^2} = \frac{{{x^2}}}{{{3^2}}}\,\,\, 
e \,\,\,\operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{x}{3}\,\,\, 
e \,\,\,\operatorname{int} \,\,\,\,\,\,\,\mathop \Rightarrow \limits^{FOCUS\,!} \,\,\,\,\,\frac{x}{3}\,\,\, 
e \,\,\,{2^3} \cdot \operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{x}{{{2^3} \cdot 3}}\,\,\, 
e \,\,\,\operatorname{int} \,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{	ext{NO}}} ightangle

The above follows the notations and rationale taught in the GMATH method.

priyaarorastep · Answer

Official Explanation: 
 Correct Answer: B

In this problem, break 24 into its prime factors: 2, 2, 2 and 3. 
Therefore, for an integer to be divisible by 24, that integer must be divisible by 2 at least three times, and 3 at least once. 
 Statement 1  
indicates that x is divisible by at least four multiples of 2 (if √xis divisible by 4, then x would be divisible by 4*4 or 2 * 2 * 2 * 2 ). 
However, statement 1 does not indicate whether x is divisible by 3, and accordingly it is not sufficient. 
 Statement 2 
 indicates that x is not divisible by 3; if xˆ2 is not divisible by 9, then the square root of x would not be divisible by the square root of 9, which is 3. As x would need to be divisible by 3 in order to be divisible by 24, it can be determined that x is not divisible by 24, and therefore the correct answer is B.

Basshead · Answer

Rephrased: Does x's prime factorization include  2^3 * 3 ?

(1) This statement tells us x is divisible by 16.

16 = 2^4 . We don't know if x is divisible by 3; INSUFFICIENT.

(2) x is not divisible by 3.

If x is not divisible by 3, we can conclude that x is not divisible by 24, since 3 is needed to satisfy the prime factorization  (2^3 * 3) . SUFFICIENT.

rvgmat12 · Answer

OE: In this problem, break 24 into its prime factors: 2, 2, 2 and 3. Therefore, for an integer to be divisible by 24, that integer must be divisible by 2 at least three times, and 3 at least once. Statement 1 indicates that x is divisible by at least four multiples of 2 (if √x is divisible by 4, then x would be divisible by 4^2, or 2 * 2 * 2 * 2 ). However, statement 1 does not indicate whether x is divisible by 3, and accordingly it is not sufficient. Statement 2 indicates that x is not divisible by 3; if x^2 is not divisible by 9, then the square root of x would not be divisible by the square root of 9, which is 3. Because x would need to be divisible by 3 in order to be divisible by 24, it can be determined that x is not divisible by 24, and therefore the correct answer is B.

Is positive integer x divisible by 24?

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