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If x and y are positive integers greater than 3, what is the remainder 06 Dec 2017, 09:05
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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.


M36-59

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If x and y are positive integers greater than 3, what is the remainder 08 Nov 2021, 09:10
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Bunuel

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.


M36-59
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Official Solution:


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.

The above means that \(|x-y|=2\).

If \(x=6\) and \(y = 4\), then the remainder when \(xy=24\) is divided by 9 is 6.

If \(x=9\) and \(y = 7\), then the remainder when \(xy=63\) is divided by 9 is 0.

Not sufficient.

(2) \(x\) and \(y\) are both primes numbers.

If \(x=y=5\), then the remainder when \(xy=25\) is divided by 9 is 7.

If \(x=y=7\), then the remainder when \(xy=47\) is divided by 9 is 4.

Not sufficient.

(1)+(2) Recall the following property:

Any prime number p, which is greater than 3, could be expressed as \(p=6n+1\) or \(p=6n-1\), where \(n\) is an integer greater than 1. For example, \(5 = 6*1 -1\), \(7 = 6*1+1\), \(11=6*2-1\), ... (Notice here that the reverse is not true. Not all number which can be expressed this way are primes. For example \(25=6*4+1\) is not a prime number.)

Since from (2) \(x\) and \(y\) are both primes numbers and from (1) one is 2 greater than the other, then we can express these primes as \(6n-1\) and \(6n+1\). In this case, \(xy=(6n-1)(6n+1)= 36n^2-1\). \(36n^2\) is divisible by 9 and -1 divided by 9 gives the remainder of 8. Sufficient.


Answer: C
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If x and y are positive integers greater than 3, what is the remainder 06 Dec 2017, 10:30
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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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If x and y are positive integers greater than 3, what is the remainder 06 Dec 2017, 10:52
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Bunuel

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.
Show more


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

Combined
the numbers would be of type 6n+1 and 6n-1..
product = \((6n+1)(6n-1) = 36n^2-1\)..
\(36n^2\) is div by 9, so remainder will be -1
but remainder has to be positive so 9-1=8
suff

C
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If x and y are positive integers greater than 3, what is the remainder 09 Dec 2017, 01:27
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Bunuel

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

Combined
the numbers would be of type 6n+1 and 6n-1..
product = \((6n+1)(6n-1) = 36n^2-1\)..
\(36n^2\) is div by 9, so remainder will be -1
but remainder has to be positive so 9-1=8
suff

C
Show more
How did you get numbers of that form 6n+1 and 6n-1?
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If x and y are positive integers greater than 3, what is the remainder 09 Dec 2017, 04:33
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Bunuel

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

Combined
the numbers would be of type 6n+1 and 6n-1..
product = \((6n+1)(6n-1) = 36n^2-1\)..
\(36n^2\) is div by 9, so remainder will be -1
but remainder has to be positive so 9-1=8
suff

C
How did you get numbers of that form 6n+1 and 6n-1?
Show more

All the PRIMES above 3 will be ODD and NOT divisible by 2 and 3
so 6n+1 and 6n-1 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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If x and y are positive integers greater than 3, what is the remainder 09 Dec 2017, 04:43
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chetan2u
stickman

How did you get numbers of that form 6n+1 and 6n-1?

All the PRIMES above 3 will be ODD and NOT divisible by 2 and 3
so 6n+1 and 6n-1 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
Show more

Perfect thanks.

So say the same qs was rephrased but instead of greater than 3, it was greater than 5, would it be 30n-1 and 30n+1?
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If x and y are positive integers greater than 3, what is the remainder 09 Dec 2017, 04:57
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How did you get numbers of that form 6n+1 and 6n-1?

All the PRIMES above 3 will be ODD and NOT divisible by 2 and 3
so 6n+1 and 6n-1 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

Perfect thanks.

So say the same qs was rephrased but instead of greater than 3, it was greater than 5, would it be 30n-1 and 30n+1?
Show more
..


No, it will always remain 6n+1 and 6n-1..

30n+1 and 30n-1 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 6n-1 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 6n-1 will be prime but PRIME will always be 6n+1 and 6n-1
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If x and y are positive integers greater than 3, what is the remainder 09 Dec 2017, 05:00
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chetan2u


No, it will always remain 6n+1 and 6n-1..

30n+1 and 30n-1 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 6n-1 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 6n-1 will be prime but PRIME will always be 6n+1 and 6n-1
Show more

I understand - thank you.
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If x and y are positive integers greater than 3, what is the remainder 09 Dec 2017, 05:07
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chetan2u


No, it will always remain 6n+1 and 6n-1..

30n+1 and 30n-1 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 6n-1 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 6n-1 will be prime but PRIME will always be 6n+1 and 6n-1

I understand - thank you.
Show more

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 \(p=6n-1\), 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 \(p=6n-1\), 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 vise-versa 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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If x and y are positive integers greater than 3, what is the remainder 24 Dec 2018, 02:50
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Bunuel

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.
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Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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If x and y are positive integers greater than 3, what is the remainder 25 Dec 2018, 22:33
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Bunuel

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.
Show more

Statement 1) The positive difference between x and y is 2. Assume x=4, y=6. The remainder (24/9)=6. Assume x=5, y=7 Remainder(35/90)=8. Two different answers. Not Sufficient.

Statement 2) x and y are both primes numbers. Any prime integer can be expressed as 6*n+1 or 6*n-1. There are three possibilities for the product of x and y. They are (6*m+1)*(6*n-1), (6*m-1)*(6m-1) or (6m+1)*(6m+1){Both expressions are the same.} and (6*n-1)*(6*n-1). Possibility 1 (6*m+1)*(6*n-1) applies to all prime numbers. Taking m=3 and n=5, so xy= 19*29. The remainder (19*29/9)=2. For m=7 and n=5, the remainder (43*29/9)=5. Two answers, Not Sufficient. Possibility 2 explores when x=y= any prime number. So, the product will be a square of the prime number. e.g. 5^2/9, the remainder is 7 or 7^2/9, the remainder is 4. Two different answers. Not Sufficient. Possibility 3 explores when the prime numbers differ by 2. So, simplifying possibility 3, (6*n-1)*(6*n-1)=(36n^2-1). Remainder when (36n^2-1) is divided by 9 is -1 or (9-1)=8. It will always be -1 or 8 irrespective of n. So, in summary, only possibility 3 of 3 possibilities gives us an accurate answer. Statement 2 on whole is Insufficient.

(1)+(2)

This is what possibility 3 discussed in 2 above assumes that the difference between x and y is 2. Hence, Sufficient.

Option C is the correct answer.
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If x and y are positive integers greater than 3, what is the remainder 24 Oct 2019, 08:29
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Bunuel

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.
Show more

\((x,y)>3; remainder(xy/9)=?\)
\(primes=[5,7,11,13,17,19,23,29,31…]\)

(1) The positive difference between x and y is 2. insufic.
\(|x-y|=2:(x,y)=(6,4)…24/9…r=6\)
\(|x-y|=2:(x,y)=(8,6)…48/9…r=3\)

(2) x and y are both primes numbers. insufic.
\((x,y)=(5,7)…35/9…r=8\)
\((x,y)=(7,11)…77/9…r=5\)

(1 & 2) sufic.
\((x,y)=(5,7)…35/9…r=8\)
\((x,y)=(11,13)…143/9=…r=8\)
\((x,y)=(29,31)…889/9=…r=8\)

Answer (C)
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If x and y are positive integers greater than 3, what is the remainder 24 Oct 2019, 08:43
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Bunuel

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.
Show more

Asked: 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.
x = 4, y = 6; xy = 24 divided by 9 gives remainder 6
x = 5; y = 7; xy = 35 divided by 9 gives remainder 8
NOT SUFFICIENT

(2) x and y are both primes numbers.
x = 7; y = 13; xy = 91 divided by 9 gives remainder 1
x = 5; y = 7; xy = 35 divided by 9 gives remainder 8
NOT SUFFICIENT

(1) + (2)
(1) The positive difference between x and y is 2.
(2) x and y are both primes numbers.
x = 5; y = 7; xy = 35 divided by 9 gives remainder 8
x = 11, y=13; xy divided by 9 gives remainder 8
x = 17; y = 19; xy divided by 9 gives remainder 8
SUFFICIENT

IMO C
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