Question

If ‘n’ and ‘a’ are positive integers such that nn = 3324 and n3 + 3n is an integral multiple of 3a, then find the largest possible value of a.

• A. 5
• B. 6
• C. 8
• D. 12
• E. 15

Answer 1

The Correct Option is D

As \(n^n =3^{324}\)
Let \(n^n =3^{324} = (3^p)^q\)
So, \(n = 3^p = q\) ..… (1)
So, \(pq = 324\)
Substituting the values of p and q such that it satisfies (1),
If \(p = 2\) and \(q = 162\) then \(3^p ≠ q\)
If \(p = 3\) and \(q = 108\) then \(3^p ≠ q\)
If \(p = 4\) and \(q = 81\) then \(3^4 = 81 = q\)
So, \(n = 81\)
Thus, \(n^3 + 3n\) \(= 81^3 +3^{81} = (34)^3 + 3^{81} = 3^{12} + 3^{81} = 3^{12}(1 + 3^{69})\)
So, the largest possible value of a is \(12\).
Hence, option D is the correct answer.