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If P and y are positive integers ; is (P^17) - 2*y^3 Odd?

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If P and y are positive integers ; is (P^17) - 2*y^3 Odd?  [#permalink]

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If p and y are positive integers, is (p^17) - 2*y^3 Odd?

A) The median of p consecutive even integers is an even integer.
B) 2p has twice as many factors as p






Do not forget to try this similar Question from MGMAT => http://gmatclub.com/forum/is-k-2-odd-118591.html

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Re: If P and y are positive integers ; is (P^17) - 2*y^3 Odd?  [#permalink]

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New post 25 Aug 2016, 21:10
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stonecold wrote:
If P and y are positive integers ; is (P^17) - 2*y^3 Odd?

A) The median of P consecutive Even integers is an Even integer.
B) 2P has twice as many factors as P






Do not forget to try this similar Question from MGMAT => is-k-2-odd-118591.html


Main question: Is (P^17) - 2*y^3 Odd?
Since 2*y^3 is always even, the question becomes -> is (P^17) odd? -> is P odd?

St1:

If P is even, median will be odd.
Example: if the set is 0,2,4,6 -> median = average of 2 and 4 which is 3.

If P is odd, median will be even.
Example: if the set is 0,2,4,6,8 -> median = 4.

Hence P is odd.
Sufficient.

St2: This is only possible if P is odd. We can try with different pairs such as (2,1) and (6,3).
Sufficient.

Answer (D).
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Re: If P and y are positive integers ; is (P^17) - 2*y^3 Odd?  [#permalink]

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New post 16 Mar 2017, 22:45
1
stonecold wrote:
If P and y are positive integers ; is (P^17) - 2*y^3 Odd?

A) The median of P consecutive Even integers is an Even integer.
B) 2P has twice as many factors as P






Do not forget to try this similar Question from MGMAT => http://gmatclub.com/forum/is-k-2-odd-118591.html


Okay..The answer is (D). I'm gonna try to explain this in detail, so that others enjoy it as much as I did it in exploring it :)

(1) The median of P consecutive Even integers is an Even integer.

Okay..consider the following series

2 4 6 8 10 12 14 16

as you can see..if you take any consecutive numbers here..one is a multiple of 4..and the other 2*(odd number)..in other words..one is an even multiple of 2 and another is an odd multiple. Now, if the number of terms are even, then the median is always the mean of the middle two numbers..say a and b are those two numbers..so it will be..
\((a+b)/2\)

which will become..
Odd + Even or Even + Odd
in either case..odd.
Thus is has to be Even number of numbers. Sufficient.

(2) 2P has twice as many factors as P

Okay..now there are two possibilities..P is either an Odd number or an Even number. In other words

\(P = a^p*b^q*c^r...\)

where a,b,c.. are all odd primes and each p,q,r.. is at least 1

or

\(P = 2^k*a^p*b^q*c^r...\) in other words

\(2^{some power}*Odd Number\)

Lets call this Odd Number = x for convenience...

Now pay attention...

If its the second case..i.e. \(P = 2^k*x\)

\(2P = 2^{k+1}*x\)

where \(x = a^p*b^q*c^r...\)

The number of factors of x will be = \((p+1)(q+1)(r+1)...\)
Let this number be = F(can be either even or odd)

Coming back to what the statement says..and putting it all into an equation...we get

\((k+2)*F = 2(k+1)*F\)

on solving..
\(k = 2k\)

but this can only be possible if k is 0. Therefore, k=0, and this is nothing but the first case :)
i.e.

\(P = a^p*b^q*c^r...\) or P is an Odd number. Sufficient.

Answer (D)
hope you enjoyed :)

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Re: If P and y are positive integers ; is (P^17) - 2*y^3 Odd?  [#permalink]

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New post 20 Feb 2018, 18:02
Can a math expert pls opine on Statement 2. My analysis was different from the other users' but I arrived at the same answer.
Bunuel chetan2u

"2P has twice as many factors as P"
I interpreted this as total # of factors, not prime factors. I reasoned that for 2P to have twice as many factors as P, P has to have 1 factor. The only positive integer that has 1 factor is 1. Thus, P is odd.
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If P and y are positive integers ; is (P^17) - 2*y^3 Odd?  [#permalink]

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New post 20 Feb 2018, 20:37
aserghe1 wrote:
Can a math expert pls opine on Statement 2. My analysis was different from the other users' but I arrived at the same answer.
Bunuel chetan2u

"2P has twice as many factors as P"
I interpreted this as total # of factors, not prime factors. I reasoned that for 2P to have twice as many factors as P, P has to have 1 factor. The only positive integer that has 1 factor is 1. Thus, P is odd.



Hi,
Your logic and reasoning are incorrect.
Every number has 1 as a factor. In other words -> 1 is the factor of every number.

If the question says factors, it is specifying total number of factors and not the prime factors.

Next, If the question mentions that 2p has twice as many factors as p => p is always odd. It can be 3 or 5 or 5001 etc. The point is that it is always odd.
You can verify it by using examples too.

Here is a GMAT-prep question based on the same logic -> https://gmatclub.com/forum/is-the-integ ... 91399.html



Best
Stone

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If P and y are positive integers ; is (P^17) - 2*y^3 Odd? &nbs [#permalink] 20 Feb 2018, 20:37
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