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Bunuel
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Solution:

The fraction is a is a positive integer. So 21^60 should be divisible by n.

21 has factors of 3,7,1,21.

So now we can eliminate A,C,D as they neither have 3 or 7 as their factors and hence cannot divide 21 or 21^60 for that matter.

E as 6 also has 3 and 2 as factor and hence they cannot create a positive integer on dividing 21 or 21^60.

So (option b)

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I went about this a bit differently, so keen to hear whether this is a 'valid' method:

As an odd number raised to any exponent is always odd, I realised \(21^{60}\) would be an odd number.

As such, it wouldn't be divisible by an even number (dividing odd by even does not result in an integer), so I immediately eliminated A, C and E - all even number solutions.

That left me with B. 3 and D. 5.

Thinking of unit digit cyclicity, I remembered any number ending in 1 raised to an exponent will end in 1... hence can't be cleanly divided by 5 as there would be a remainder.

By process of elimination, the answer seems to be B. 3
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\(21^{60}\) will yield '1' at the unit place.

Then, only '3' can be the answer from given choices as other numbers have no multiples ending in '1'.

Answer B
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Bunuel
If \(\frac{21^{60}}{n}\) is a positive integer, which of the following could be n?

A. 2
B. 3
C. 4
D. 5
E. 6

\(\frac{21^{60}}{n}\) = \(\frac{3^{60}*7^{60}}{n}\)

i.e. The number is divisible by 3 and 7 both

A. 2 is Not a factor of given number
B. 3 is a factor of given number
C. 4 is Not a factor of given number
D. 5 is Not a factor of given number
E. 6 is Not a factor of given number


Answer: Option B
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