Amateur wrote:

This is a quant question from one of Manhattan's Flash cards

For each of the following, could the answer be an integer if x is an integer greater than 1?

b) x^1/6 + x^1/2 =

His explanation is

Yes. This is equivalent to 6th \sqrt{x} x + \sqrt{x} , so if x has an integer

sixth root this will be an integer. For example, if x equals 64, the sixth root of x is 2, and the square root is 8.

question was if x is integer greater than 1. What if 6th root is not an integer.... sum won't be an integer right? and in explanation he uses "will be an integer" (in bold) how is it possible?

Well the question states that can the sum be an integer not must. In other words, I think the question is asking, is it possible for the given expression to have integer values for integer values of x. So we have to establish cases where the value can be an integer. The basic explanation would be,

if x = m^6 where m is an integer, then x^(1/6) = m

also, x^(1/2) = m^3. Hence there will always be an m for which the above expression will be an integer. Hence m belongs to the soln set {2^6,3^6,.....} for which the expression can be an integer.

So the explanation provided, that the sixth root needs to be an integer is accurate, as if the value x^1/6 is not an integer, the expression wont be.

Hope the solution is satisfactory

Please correct me if I am wrong!

Regards,

Arpan

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