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As the OA is not provided can someone please let me know whether my solution is correct or not?

Considering question stem

Cannot be simplified any further apart from prime factorization for 24 which are 2^3 * 3

Considering Statement 1

n is ODD.

When n =1 Remainder will be zero. n=3 Remainder won't be zero n =5 Remainder will be zero Two answers therefore insufficient.

Considering Statement 2

n is not a multiple of 3. As it will give different value of r this statement alone is insufficient.

Combining both statement 1 & 2

n is not a multiple of 6 i.e. 2 and 3. So n is prime without 2. Therefore n can be 5, 7, 11,..etc and the remainder will be ZERO.Therefore answer should be c i.e. both statements together are sufficient to answer the question.

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r? 1). n is not divisible by 2 2). n is not divisible by 3

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

Plug-in method:

\((n-1)(n+1)=n^2-1\)

(1) n is not divisible by 2 --> pick two odd numbers: let's say 1 and 3 --> if \(n=1\), then \(n^2-1=0\) and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if \(n=3\), then \(n^2-1=8\) and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.

(2) n is not divisible by 3 --> pick two numbers which are not divisible by 3: let's say 1 and 2 --> if \(n=1\), then \(n^2-1=0\), so remainder is 0 but if \(n=2\), then \(n^2-1=3\) and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.

(1)+(2) Let's check for several numbers which are not divisible by 2 or 3: \(n=1\) --> \(n^2-1=0\) --> remainder 0; \(n=5\) --> \(n^2-1=24\) --> remainder 0; \(n=7\) --> \(n^2-1=48\) --> remainder 0; \(n=11\) --> \(n^2-1=120\) --> remainder 0. Well it seems that all appropriate numbers will give remainder of 0. Sufficient.

Algebraic approach:

(1) n is not divisible by 2. Insufficient on its own, but this statement says that \(n=odd\) --> \(n-1\) and \(n+1\) are consecutive even integers --> \((n-1)(n+1)\) must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).

(2) n is not divisible by 3. Insufficient on its own, but form this statement either \(n-1\) or \(n+1\) must be divisible by 3 (as \(n-1\), \(n\), and \(n+1\) are consecutive integers, so one of them must be divisible by 3, we are told that it's not \(n\), hence either \(n-1\) or \(n+1\)).

(1)+(2) From (1) \((n-1)(n+1)\) is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by \(8*3=24\), which means that remainder upon division \((n-1)(n+1)\) by 24 will be 0. Sufficient.

Re: What is the remainder? (n-1)(n+1) divided by 24 [#permalink]

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31 Dec 2012, 20:18

5

This post received KUDOS

Nadezda wrote:

If n is positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

(1) is not divisible by 2 (2) is not divisible by 3

Statement 1) When n is not divisible by 2, then n can be \(1, 3, 5, 7, 9 etc\) For n=1, the remainder is 0 For n=3, the remainder is 16. For n=5, the remainder is 0. Different answers. Hence insufficient.

statement 2) When n is not divisible by 3, then n can be \(1,2, 4, 6 etc\) Here also different remainders. Insufficient.

On combining these two statements, n is \(1,5, 7 etc\) For such numbers, the remainder is 0. Sufficient. +1C
_________________

Re: If n is a positive integer and r is the remainder when [#permalink]

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12 Jan 2013, 02:34

1

This post received KUDOS

kiyo0610 wrote:

If n is a positive integer and r is the remainder when (n-1)(n+1)is divided by 24, what is the value of r?

(1) n is not divisible by 2 (2) n is not divisible by 3

n-1,n, n+1 are consecutive +ve intigers, and thus if n is even both n-1,n+1 are odd and vice versa. also in every 3 consecutive numbers we get one that is a multiple of 3

from 1

n is odd thus both n-1, n+1 are even and their product has at least 2^3 as a factor however if n = 3 thus n-1,n+1 are 2,4 and since , 24 = 2^3*3 , thus reminder is 3 but if n = 5 for example thus n-1,n+1 are 4,6 and therofre in this case r = 0.....insuff

from 2

n is a multiple of 3 and thus both n-1,n+1 are either even (e.g: n=3) or odd (n=6) and therfore this is insuff

both together

n is odd and is a multiple of 3 and therfore the reminder of the product (n-1)(n+1) when devided by 24 is always 3..suff

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r? a)n is not divisible by 2 b) n is not divisible by 3

A few things: 1. Exactly one of any two consecutive positive integers is even. 2. Exactly one of any three consecutive positive integers must be a multiple of 3 3. Exactly one of any four consecutive positive integers must be a multiple of 4 etc Check this post for the explanation: http://www.veritasprep.com/blog/2011/09 ... c-or-math/

a) n is not divisible by 2

Since every alternate number is divisible by 2, (n-1) and (n+1) both must be divisible by 2. Since every second multiple of 2 is divisible by 4, one of (n-1) and (n+1) must be divisible by 4. Hence, the product (n-1)*(n+1) must be divisible by 8. But if n is divisible by 3, then neither (n-1) nor (n+1) will be divisible by 3 and hence, when you divide (n-1)(n+1) by 24, you will get some remainder. If n is not divisible by 3, one of (n-1) and (n+1) must be divisible by 3 and hence the product (n-1)(n+1) will be divisible by 24 and the remainder will be 0. Not sufficient.

b) n is not divisible by 3 We don't know whether n is divisible by 2 or not. As discussed above, we need to know that to figure whether the product (n-1)(n+1) is divisible by 8. Hence not sufficient.

Take both together, we know that (n-1)*(n+1) is divisible by 8 and one of (n-1) and (n+1) is divisible by 3. Hence, the product must be divisible by 8*3 = 24. So r must be 0. Sufficient.

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r? a)n is not divisible by 2 b) n is not divisible by 3

Given:

(n-1)(n+1) = 24m + r - (1) where m=1,2,3...

Statement 1:

n = 2w + x - (2)

Statement 1 alone is not sufficient

Statement 2:

n= 3y + z - (3)

Statement 2 alone is not sufficient.

Taken together:

1. x has to be 1 and z can be 1 or 2

2. When x and z are 1, the values of n are 7, 13, 19 etc

3. When x=1 and z=2, the values of n are 5, 11, 17 etc

4. Substitute one of the above values, say 5 in (1)

5. For n=5 we have 4*6 = 24m + r or 24m +r = 24 r=0

We will get the same value of r for the other values of n too.

Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]

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13 Jan 2015, 14:00

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Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]

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26 Jan 2016, 18:33

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Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]

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24 Oct 2016, 07:08

Bunuel wrote:

enigma123 wrote:

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r? 1). n is not divisible by 2 2). n is not divisible by 3

If n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24, what is the value of r?

Plug-in method:

\((n-1)(n+1)=n^2-1\)

(1) n is not divisible by 2 --> pick two odd numbers: let's say 1 and 3 --> if \(n=1\), then \(n^2-1=0\) and as zero is divisible by 24 (zero is divisible by any integer except zero itself) so remainder is 0 but if \(n=3\), then \(n^2-1=8\) and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.

(2) n is not divisible by 3 --> pick two numbers which are not divisible by 3: let's say 1 and 2 --> if \(n=1\), then \(n^2-1=0\), so remainder is 0 but if \(n=2\), then \(n^2-1=3\) and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.

(1)+(2) Let's check for several numbers which are not divisible by 2 or 3: \(n=1\) --> \(n^2-1=0\) --> remainder 0; \(n=5\) --> \(n^2-1=24\) --> remainder 0; \(n=7\) --> \(n^2-1=48\) --> remainder 0; \(n=11\) --> \(n^2-1=120\) --> remainder 0. Well it seems that all appropriate numbers will give remainder of 0. Sufficient.

Algebraic approach:

(1) n is not divisible by 2. Insufficient on its own, but this statement says that \(n=odd\) --> \(n-1\) and \(n+1\) are consecutive even integers --> \((n-1)(n+1)\) must be divisible by 8 (as both multiples are even and one of them will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8, 10); (10, 12), ...).

(2) n is not divisible by 3. Insufficient on its own, but form this statement either \(n-1\) or \(n+1\) must be divisible by 3 (as \(n-1\), \(n\), and \(n+1\) are consecutive integers, so one of them must be divisible by 3, we are told that it's not \(n\), hence either \(n-1\) or \(n+1\)).

(1)+(2) From (1) \((n-1)(n+1)\) is divisible by 8, from (2) it's also divisible by 3, therefore it must be divisible by \(8*3=24\), which means that remainder upon division \((n-1)(n+1)\) by 24 will be 0. Sufficient.

Answer: C.

Hope it's clear.

Shouldn't it be numbers which are not divisible by 2 and 3?

gmatclubot

Re: If n is a positive integer and r is the remainder when (n-1)
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24 Oct 2016, 07:08

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