If f(n) denotes the remainder of n^2 divided by k, where n and k are

Sol: IMO-Option-E
Tried to answer it with a different method, but it doesn't match the OA though.


\(f(n)\) denotes the remainder of \(n^2\) divided by k == > \(n^2\) = kQ + \(f(n)\) , where Quotient, Q > 0
or \(f(n)\) = \(n^2\) - kQ

St-1:
\(f(k+12) = 16\)
==> 16 = (k+12)^2 - kQ
L.H.S = even
Depending whether k & Q are odd or even four cases arise:
Case-1:
k= Odd , Q = Odd
R.H.S = Odd - Odd*Odd = Even = L.H.S ---> Valid Case
Case-2:
k = Odd , Q=Even
R.H.S = Odd - Odd*Even= Odd ---> Invalid Case
Case-3:
k= Even , Q = Odd
R.H.S = Even- Even*Odd = Odd ---> Invalid Case
Case-4:
k = Even; Q=Even
R.H.S = Even - Even*Even= Odd = L.H.S ---> Valid Case
We have two valid cases, in first k is odd and in another k is even. Statement is Insufficient

St-2:
\(f(k+11) = 9\)
==> 9 = (k+11)^2 - kQ
L.H.S = Odd
Depending whether k & Q are odd or even four cases arise:
Case-1:
k= Odd , Q = Odd
R.H.S = Even - Odd*Odd = Odd = L.H.S ---> Valid Case
Case-2:
k = Odd , Q=Even
R.H.S = Even - Odd*Even= Even ---> Invalid Case
Case-3:
k= Even , Q = Odd
R.H.S = Odd - Even*Odd = Odd ---> Valid Case
Case-4:
k = Even; Q=Even
R.H.S = Odd - Even*Even= Odd = L.H.S ---> Valid Case
We have three valid cases, in Case-1 k is odd and in Case 3 and 4 k is even. Statement is Insufficient

Combining St-1 & 2

Case-1 and 4 are still common , so k can be even or odd. Both statements are together not sufficient

Bunuel is there a remainder based approach ?l