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13 Jan 2014, 01:03
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D
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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.
(1) x^2 is divisible by 9.
(2) x^4 is divisible by 9.
13 Jan 2014, 01:13
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If x is an integer, is x^3 divisible by 9?
Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?
(1) x^2 is divisible by 9 --> x^2 is divisible by 3^2 --> x is divisible by 3 --> x^3 is divisible by 3^3*k, thus it's divisible by 9 too. Sufficient.
(2) x^4 is divisible by 9 --> the same here: x^4 is divisible by 3^2 --> x is divisible by 3 --> x^3 is divisible by 3^3*k, thus it's divisible by 9 too. Sufficient.
Answer: D.
Hope it's clear.
Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?
(1) x^2 is divisible by 9 --> x^2 is divisible by 3^2 --> x is divisible by 3 --> x^3 is divisible by 3^3*k, thus it's divisible by 9 too. Sufficient.
(2) x^4 is divisible by 9 --> the same here: x^4 is divisible by 3^2 --> x is divisible by 3 --> x^3 is divisible by 3^3*k, thus it's divisible by 9 too. Sufficient.
Answer: D.
Hope it's clear.
General Discussion
01 Feb 2014, 07:47
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Hi Bunuel, is it okay to reason stmt2 as below?
\(x^4\) is divisible by 9.
then either
a. x is divisible by 9 or
b. \(x^3\) is divisible by 9.
If a is true then b is also true.
if a is not true then b has to be true. therefore for \(x^4\) to be divisible by 9, \(x^3\) MUST be divisible by 9.
\(x^4\) is divisible by 9.
then either
a. x is divisible by 9 or
b. \(x^3\) is divisible by 9.
If a is true then b is also true.
if a is not true then b has to be true. therefore for \(x^4\) to be divisible by 9, \(x^3\) MUST be divisible by 9.
