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

Answer: (D)

12

Answer 2

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.