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If p, m, and n are positive integers, n is odd, and p = m^2

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If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) m – n = 3

Originally posted by maniac on 24 Feb 2014, 08:44.
Last edited by Bunuel on 24 Feb 2014, 08:58, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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New post 24 Feb 2014, 11:36
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Option A.
From S1:p=8k +5
Implies p=odd
n^2=odd
p=m^2+n^2
m^2=even; implies m=even
Since all are +ve integers m=2,4,6,...
Now condition is p=m^2+n^2
Smallest value of p=13
Possible values for m,n are
M=2;n=3
Implies m isnt div by 4.
If p=21 m and n are not possible since relationship doesnt hold.
If p=29 m=2;n=5 again m isnt div by 4
If p=37 m=6;n=1 again m isnt div by 4.
Got convinced at this pt. that m wont be div by 4.

From S2:m-n=3
Squaring m^2+n^2=9+2mn
p=9+2mn
P=odd;n=odd.
But nothing about m.
Not suff.

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Re: If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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New post 01 Jul 2014, 03:12
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Statement 1

p=m^2+n^2
R of (p/8)=R of [m^2/8]+ R of [n^2/8]
1=R of [m^2/8]+ R of [(odd integer square)/8]
5=R of [m^2/8]+1
R of [m^2/8]=4
m^2=8x+4
m^2=4(2x+1)↪2x+1 is an odd square
m=2.odd integer…..not divisible by 4…sufficient

Statement 2

m=n+3↪↪↪whic h is three more than an odd integer↪↪could be 9+3=12 or 11+3=14..not sufficient

Answer-A
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Re: If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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New post 01 Jul 2014, 21:06
Bunuel can we have some of your expert analysis. I am a bit confused on the analysis given by AKG1593. He put in values to a certain number and postulated that it will be true for ALL numbers. How do we know that such generalization across entire number set is true ?
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Re: If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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New post 01 Jul 2014, 21:25
1
maniac wrote:
If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) m – n = 3


Solution:

Given that p, m and n are positive integers. If n is odd, n^2 must also be odd. How do you represent an odd integer? As (2k + 1)

So n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1

Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. Therefore, when n^2 is divided by 8, it will leave a remainder of 1.

Statement 1: When p is divided by 8, the remainder is 5.

When p ( i.e. m^2 + n^2) is divided by 8, we get a remainder of 5. When n^2 is divided by 8, the remainder will be 1. To get a remainder of 5, when m^2 is divided by 8, we should get a remainder of 4.
m^2 = 8a + 4 (i.e. we can make ‘a’ groups of 8 and 4 will be leftover)
m^2 = 4(2a + 1)
This implies m = 2*√(Odd Number) because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that m is not divisible by 4.
This statement alone is sufficient.

Statement 2: m – n = 3

The difference between m and n is 3 i.e. an odd number. Since n is odd, we can say that m will be even (Even – Odd = Odd). But whether m is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p.
This statement alone is not sufficient.

Answer (A)

This solution and this concept is discussed here: http://www.veritasprep.com/blog/2013/01 ... roperties/
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Re: If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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New post 24 Sep 2014, 00:21
It should be sufficient to prove that m is even since that means that m is a multiple of 2 and m^2 will be a multiple of 4.

Since n is odd and stm(1) says p = 8k + 5 (i.e p is also odd), m^2 has to be even which in turn implies m is even.
p = m^2 + n^2
odd = even + odd

Hence the answer is (A)
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Re: If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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New post 24 Sep 2014, 10:31
maniac wrote:
If p, m, and n are positive integers, n is odd, and p = m^2 + n^2, is m divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) m – n = 3


A.

1) p = 8k+5
so p is odd.
also, p = m^2 + n^2
n is odd so for p to be odd m needs to be even.
1^2 mod 8 = 1
2^2 mod 8 = 4
3^2 mod 8 = 1
4^2 mod 8 = 0
and then it repeats.
we know (m^2 + n^2) mod 8 = 5 ; where m is even and n is odd
in the range given above m can be 2 and n can be either 1 or 3
m and n both have a cyclicity of 4 --> so m would be 2,6,10,....(4n+2 form) which is never divisible by 4.
so 1 alone is sufficient.

2) m-n=3
so m is even.
even means 2k and not necessarily 4k. hence 2 alone is inconclusive.
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If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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New post 10 Aug 2016, 07:51
Is this really a GMAT like question
I really doubt that.

Isnt it a poor quality question . you know , similar to the ones i write :)



It took me 4 and a half minutes to solve this

Any shortcuts Vyshak Abhishek009

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Re: If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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New post 13 Aug 2016, 12:12
Was able to solve it in 3:30mins.

since n is odd; it can be written as : (2k+1)
Stmt1: p can be expressed as : 8k+5
putting this in original question, we get: m^2 = 8k+5 - (2a+1)^2
m^2 = 8k + 5 - 4a^2 - 1 - 4a
m^2 = 8k + 4 - 4a^2 - 4a
m^2 = 4 x (...)
therefore, m is divisible by 4. Sufficient.

Stmt2: Not sufficient.

A.
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Re: If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]

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Re: If p, m, and n are positive integers, n is odd, and p = m^2   [#permalink] 13 Apr 2018, 09:03
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