Is positive integer x divisible by 24?

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.

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

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

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

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

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

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

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

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.

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.

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.