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

Question

If  f(n)  denotes the remainder of  n^2  divided by k, where n and k are positive integers, is k an even number?

(1)  f(k+12) = 16 
(2)  f(k+11) = 9

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For positive integers n and k, if  f (n)  represents the remainder when  n^2  is divided by k, is k an even number?

1)  f(k+12)=16 
2)  f(k+11)=9

Aryaa03 · Answer

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

wytwyt · Answer

The solution seems wrong

Take Statement 1: f(k + 12) is (k+12)^2 mod k = (k*k + 2*k*12 + 144) mod k = 16
Meaning 144 mod k = 16 => k divides 128 and k is a factor of 128 and 128 has only even factors. So k must be even

Aryaa03 · Answer

Yes, I am sure about the incorrectness, as the OA is released to be D.
Can someone explain the step where it went wrong or started wrong?

Posted from my mobile device

Bunuel · Answer

If  f(n)  denotes the remainder of  n^2  divided by k, where n and k are positive integers, is k an even number?

(1)  f(k+12) = 16

The above implies that (k + 12)^2 = k^2 + 24k + 144 divided by k yields a remainder of 16 and that k > 16 (the divisor must be greater than the remainder). Since both k^2 and 24k are divisible by k, then 144 divided by k yields a remainder of 16: 144 = kq + 16. This gives kq = 128, so k must be a factor of 128 and greater than 16. Since all factors of 128 greater than 16 are even, k is even. Sufficient.

(2)  f(k+11) = 9

The above implies that (k + 11)^2 = k^2 + 22k + 121 divided by k yields a remainder of 9 and that k > 9 (the divisor must be greater than the remainder). Since both k^2 and 22k are divisible by k, then 121 divided by k yields a remainder of 9: 121 = kp + 9. This gives kp = 112, so k must be a factor of 112 and greater than 9. Since all factors of 112 greater than 9 are even, k is even. Sufficient.

Answer: D.

Hope it helps.

bumpbot · Answer

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

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

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