If x is an integer, is x^3 divisible by 9?
(1) x^2 is divisible by 9.
(2) x^4 is divisible by 9.
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MartyMurray
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Statement 1: x² is divisible by 9.
x³ = x² * x
So if x² is divisible by 9, then x³ is divisible by 9.
Sufficient.
Statement 2: x� is divisible by 9.
9 = 3 * 3
So if x� is divisible by 9, then among the prime factors of x� are at least two 3's.
From the question we know that x is an integer. Meanwhile, 3 is prime.
If x� is divisible by 9, then x� = 3 * 3 * (some other integer). To get an integer x, the 4th root of the right side of the equation has to be an integer.
In order for the 4th root of the right side to be an integer, we need four sets of the same prime factors. So we need a total of four 3's among the prime factors of x�.
So, since x is an integer and 3 is prime, if there are among the prime factors of x� two 3's, there must be four 3's.
So x� = 3� * (some other integer)�.
Therefore, x³ = 3³ * (some other integer)³.
3³ is divisible by 9. So x³ is divisible by 9.
Sufficient.
The correct answer is D.
x³ = x² * x
So if x² is divisible by 9, then x³ is divisible by 9.
Sufficient.
Statement 2: x� is divisible by 9.
9 = 3 * 3
So if x� is divisible by 9, then among the prime factors of x� are at least two 3's.
From the question we know that x is an integer. Meanwhile, 3 is prime.
If x� is divisible by 9, then x� = 3 * 3 * (some other integer). To get an integer x, the 4th root of the right side of the equation has to be an integer.
In order for the 4th root of the right side to be an integer, we need four sets of the same prime factors. So we need a total of four 3's among the prime factors of x�.
So, since x is an integer and 3 is prime, if there are among the prime factors of x� two 3's, there must be four 3's.
So x� = 3� * (some other integer)�.
Therefore, x³ = 3³ * (some other integer)³.
3³ is divisible by 9. So x³ is divisible by 9.
Sufficient.
The correct answer is D.
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Jay@ManhattanReview
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We know that 9 = 3^2; for X^3 to be divisible by 9, two 3s should be canceled with the prime factors of X^3.prith24 wrote:If x is an integer, is x^3 divisible by 9?
(1) x^2 is divisible by 9.
(2) x^4 is divisible by 9.
S1: Since X^2 is divisible by 9 = 3^2, X is a multiple of 3. Sufficient!
S2: Though if X were a square root of 3, the condition: X^4 is divisible by 9 would hold true, and then X^3 will NOT be divisible by 9. However, we are given that X is an integer, thus, X is a multiple of 3. Sufficient!
OA: D
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Jeff@TargetTestPrep
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We are given that x is an integer and need to determine whether (x^3)/9 = integer.prith24 wrote:If x is an integer, is x^3 divisible by 9?
(1) x^2 is divisible by 9.
(2) x^4 is divisible by 9.
Statement One Alone:
x^2 is divisible by 9.
Since 9 = 3^2, (x^2)/9 = integer means x/3 = integer (notice that (x^2)/9 = (x^2)/(3^2) = (x/3)^2). Therefore, we know that x is a multiple of 3. Since x is a multiple of 3, any multiple of 3 raised to the 3rd power will always be divisible by 9. Statement one alone is sufficient to answer the question. We can eliminate answer choices B, C, and E.
Statement Two Alone:
x^4 is divisible by 9.
Again, since 9 = 3^2, (x^4)/9 = integer means (x^2)/3 = integer. Since 3 is a prime number and x is an integer, (x^2)/3 = integer, and thus x/3 = integer. Therefore, we know that x is a multiple of 3.
Since x is a multiple of 3, any multiple of 3 raised to the 4th power will always be divisible by 9. Statement two alone is sufficient to answer the question.
Answer: D
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