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If x and y are positive integers greater than 3, what is the remainder [#permalink]
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06 Dec 2017, 08:05
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Re: If x and y are positive integers greater than 3, what is the remainder [#permalink]
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06 Dec 2017, 09:30
Imo C St1: X=9, y=7 Remainder = 0 X=6, y= 4 Remainder = 6 Not sufficient St2: X=5 Y=7 Remainder = 8 X=5 Y= 5 Remainder = 7 Not sufficient Combining 1 and 2 X=9, y=7 X=13, y=11 X= 19, y= 17 All gives same remainder 8 Sufficient Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app



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Re: If x and y are positive integers greater than 3, what is the remainder [#permalink]
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06 Dec 2017, 09:52
Bunuel wrote: GMAT Club's Fresh Challenge Problem. If x and y are positive integers greater than 3, what is the remainder when xy is divided by 9? (1) The positive difference between x and y is 2. (2) x and y are both primes numbers. a proper method would be (1) The positive difference between x and y is 2.many cases possible.. 9 and 7, remainder is 0 7 and 5 , remainder is 8 insuff (2) x and y are both primes numbers.5 and 7 will give 8 as remainder 7 and 11 will give 5 as remainder Insuff Combinedthe numbers would be of type 6n+1 and 6n1.. product = \((6n+1)(6n1) = 36n^21\).. \(36n^2\) is div by 9, so remainder will be 1 but remainder has to be positive so 91=8 suff C
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Re: If x and y are positive integers greater than 3, what is the remainder [#permalink]
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09 Dec 2017, 00:27
chetan2u wrote: Bunuel wrote: GMAT Club's Fresh Challenge Problem. If x and y are positive integers greater than 3, what is the remainder when xy is divided by 9? (1) The positive difference between x and y is 2. (2) x and y are both primes numbers. a proper method would be (1) The positive difference between x and y is 2.many cases possible.. 9 and 7, remainder is 0 7 and 5 , remainder is 8 insuff (2) x and y are both primes numbers.5 and 7 will give 8 as remainder 7 and 11 will give 5 as remainder Insuff Combinedthe numbers would be of type 6n+1 and 6n1.. product = \((6n+1)(6n1) = 36n^21\).. \(36n^2\) is div by 9, so remainder will be 1 but remainder has to be positive so 91=8 suff C How did you get numbers of that form 6n+1 and 6n1?



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Re: If x and y are positive integers greater than 3, what is the remainder [#permalink]
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09 Dec 2017, 03:33
stickman wrote: chetan2u wrote: Bunuel wrote: GMAT Club's Fresh Challenge Problem. If x and y are positive integers greater than 3, what is the remainder when xy is divided by 9? (1) The positive difference between x and y is 2. (2) x and y are both primes numbers. a proper method would be (1) The positive difference between x and y is 2.many cases possible.. 9 and 7, remainder is 0 7 and 5 , remainder is 8 insuff (2) x and y are both primes numbers.5 and 7 will give 8 as remainder 7 and 11 will give 5 as remainder Insuff Combinedthe numbers would be of type 6n+1 and 6n1.. product = \((6n+1)(6n1) = 36n^21\).. \(36n^2\) is div by 9, so remainder will be 1 but remainder has to be positive so 91=8 suff C How did you get numbers of that form 6n+1 and 6n1? All the PRIMES above 3 will be ODD and NOT divisible by 2 and 3 so 6n+1 and 6n1 will give you all numbers not div by 2 and 3 AND only two consecutive odd numbers will be prime as the THIRD will be surely be multiple of 3.. example 5,7, 9........ 9,11,13.....11,13, 15
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Re: If x and y are positive integers greater than 3, what is the remainder [#permalink]
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09 Dec 2017, 03:43
chetan2u wrote: stickman wrote: How did you get numbers of that form 6n+1 and 6n1? All the PRIMES above 3 will be ODD and NOT divisible by 2 and 3 so 6n+1 and 6n1 will give you all numbers not div by 2 and 3 AND only two consecutive odd numbers will be prime as the THIRD will be surely be multiple of 3.. example 5,7, 9........ 9,11,13.....11,13, 15Perfect thanks. So say the same qs was rephrased but instead of greater than 3, it was greater than 5, would it be 30n1 and 30n+1?



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If x and y are positive integers greater than 3, what is the remainder [#permalink]
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09 Dec 2017, 03:57
stickman wrote: chetan2u wrote: stickman wrote: How did you get numbers of that form 6n+1 and 6n1? All the PRIMES above 3 will be ODD and NOT divisible by 2 and 3 so 6n+1 and 6n1 will give you all numbers not div by 2 and 3 AND only two consecutive odd numbers will be prime as the THIRD will be surely be multiple of 3.. example 5,7, 9........ 9,11,13.....11,13, 15Perfect thanks. So say the same qs was rephrased but instead of greater than 3, it was greater than 5, would it be 30n1 and 30n+1? .. No, it will always remain 6n+1 and 6n1.. 30n+1 and 30n1 will miss many prime numbers .. if n =1 , the number becomes 31 and 29, both are prime but it misses out on many other prime like 7,11,13.. reason we take 6n+1 and 6n1 is that MOST of the numbers are div by 2 and 3 and therefore, possibility of missing any prime is 0.. n=1.....7 and 5 n=2.....13 and 11 n=3.....19 and 17 so here we are avoiding EVEN numbers and multiples of 3  9,15,21 and so on.. BUT remember it is not necessary that 6n+1 and 6n1 will be prime but PRIME will always be 6n+1 and 6n1
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If x and y are positive integers greater than 3, what is the remainder [#permalink]
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09 Dec 2017, 04:00
chetan2u wrote: No, it will always remain 6n+1 and 6n1..
30n+1 and 30n1 will miss many prime numbers .. if n =1 , the number becomes 31 and 29, both are prime but it misses out on many other prime like 7,11,13..
reason we take 6n+1 and 6n1 is that MOST of the numbers are div by 2 and 3 and therefore, possibility of missing any prime is 0.. n=1.....7 and 5 n=2.....13 and 11 n=3.....19 and 17
so here we are avoiding EVEN numbers and multiples of 3  9,15,21 and so on..
BUT remember it is not necessary that 6n+1 and 6n1 will be prime but PRIME will always be 6n+1 and 6n1
I understand  thank you.



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Re: If x and y are positive integers greater than 3, what is the remainder [#permalink]
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09 Dec 2017, 04:07
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stickman wrote: chetan2u wrote: No, it will always remain 6n+1 and 6n1..
30n+1 and 30n1 will miss many prime numbers .. if n =1 , the number becomes 31 and 29, both are prime but it misses out on many other prime like 7,11,13..
reason we take 6n+1 and 6n1 is that MOST of the numbers are div by 2 and 3 and therefore, possibility of missing any prime is 0.. n=1.....7 and 5 n=2.....13 and 11 n=3.....19 and 17
so here we are avoiding EVEN numbers and multiples of 3  9,15,21 and so on..
BUT remember it is not necessary that 6n+1 and 6n1 will be prime but PRIME will always be 6n+1 and 6n1
I understand  thank you. chetan2u is referring to the following property: Any prime number p, which is greater than 3, could be expressed as \(p=6n+1\) or \(p=6n+5\) or \(p=6n1\), where n is an integer greater than 1.Any prime number p, which is greater than 3, when divided by 6 can only give the remainder of 1 or 5 (remainder cannot be 2 or 4 as in this case p would be even and the remainder cannot be 3 as in this case p would be divisible by 3). So, any prime number p, which is greater than 3, could be expressed as \(p=6n+1\) or \(p=6n+5\) or \(p=6n1\), where n is an integer greater than 1. But:Not all number which yield a remainder of 1 or 5 upon division by 6 are primes, so viseversa of the above property is not true. For example 25 yields the remainder of 1 upon division be 6 and it's not a prime number. Hope it's clear.
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Re: If x and y are positive integers greater than 3, what is the remainder
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