A positive integer K leaves no remainder when divided by an integer Y

Question

A positive integer K leaves no remainder when divided by an integer Y and the quotient is Q. Is Q odd, if PK^2 + Z is even, where Z and P are integers?

(1) Y is odd and Z is even
(2) PK^2 and K^3 have equal number of distinct prime factors

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chetan2u · Answer

PK^2+Z  is even.
So either both PK and Z are even or both are odd.
Also, K=YQ

(1) Y is odd and Z is even
 PK^2+Z=PK^2+even=even .  So PK^2 is even . => P(YQ)^2=even
P(odd*Q)=even........PQ is even
But we cannot say anything clearly about Q.

(2) PK^2 and K^3 have equal number of distinct prime factors
K is common in both, so P also has same or lesser distinct prime factors.
 If K is odd, P is surely odd.  But if K is even, P could be anything.
Nothing about Q

Combined,
PK^2 is even, and if K is odd, P is surely odd.
If K is odd, then PK^2=odd*(odd)^2=odd
eq{even}.
So K is surely even.=> YQ=even.....odd*Q=even. So Q is surely even.
Sufficient

C

Pankaj0901 · Answer

Answer IMO C

S1: 
 P could be Even or Odd, hence cannot say about Z. 
 NOT SUFFICIENT

S2: 
 PK^2 and K^3 have same distinct Prime factors
This implies, either P is a factor of K, or K is a factor of P. Or, P=1
Example 1: K=25 and P=5
PK^2 and K^3 both have 5 has the Prime factor.
Example 2: K=25 and P=625
Same as example 1..

This implies the nature of P will be same as nature of K, except that P can be equal to 1.
But it doesn't tell about the nature of Z or K itself (as S1 does)
Hence,  NOT SUFFICIENT

Now, if combined with S1: we can identify the nature of P (as we know about K) and thus determine whether the expression PK^3 + Z is EVEN or ODD.

=>  BOTH TOGETHER ARE SUFFICIENT

bumpbot · Answer

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A positive integer K leaves no remainder when divided by an integer Y

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