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Bunuel
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Hi Bunuel, sorry but how did you infer that x, and y must be perfect cubes - didnt connect the two while solving? thank you!

To form a larger cube using small identical cubes, you need an n by n by n arrangement. So if the number of small cubes used is x, then x = n^3 for some integer n. That’s why x and y must be perfect cubes. You can’t form a cube with, say, 10 unit cubes, only numbers like 1, 8, 27, 64, etc. (which are perfect cubes).
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Bunuel, Bunuel,

Couldnt there be an extra variable between your x and your n in x = n^3

I mean could it be x=p n^3

or does n takes care of everything ( each individual smaller side* number of smaller cubes) and this must be a cube itself?

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Juan C. Avellan


Bunuel

To form a larger cube using small identical cubes, you need an n by n by n arrangement. So if the number of small cubes used is x, then x = n^3 for some integer n. That’s why x and y must be perfect cubes. You can’t form a cube with, say, 10 unit cubes, only numbers like 1, 8, 27, 64, etc. (which are perfect cubes).
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Bunuel, Bunuel,

Couldnt there be an extra variable between your x and your n in x = n^3

I mean could it be x=p n^3

or does n takes care of everything ( each individual smaller side* number of smaller cubes) and this must be a cube itself?

Regards,

Juan C. Avellan




No, x itself must be a perfect cube. If the big cube is n by n by n, then the total number of small cubes is exactly n^3. So x = n^3, nothing extra.
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I don’t quite agree with the solution. x & y need not be cubes right?
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I don’t quite agree with the solution. x & y need not be cubes right?
Check here: https://gmatclub.com/forum/m37-375682.html#p3554662 Hope it helps.
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