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If p, m, and n are positive integers, n is odd, and p = m^2 [#permalink]
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Updated on: 24 Feb 2014, 08:58
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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
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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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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:mn=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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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
AnswerA



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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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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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01 Jul 2014, 21:25
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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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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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) mn=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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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 Abhishek009Regards Stone Cold Steve Austin
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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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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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13 Apr 2018, 09:03
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