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If n is a positive integer and r is the remainder when (n - 1)(n + 1)

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

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New post 21 Jan 2012, 17:57
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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

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

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New post 21 Jan 2012, 18:12
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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.
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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1)  [#permalink]

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New post 31 Dec 2012, 21:18
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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.
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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1)  [#permalink]

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New post 25 Mar 2013, 21:02
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vbodduluri 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?
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.

Answer (C)

Another approach:

(n-1), n and (n+1) are consecutive integers.

If 2 is not a factor of n (i.e. n is odd), it must be a factor of (n-1) and (n+1) (the numbers around n must be even).
Also, out of any two consecutive even numbers, one has to be divisible by 4 because every alternate multiple of 2 is divisible by 4.
Hence, (n-1)*(n+1) must be divisible by 8.

Out of any 3 consecutive integers, one has to be divisible by 3 because every third integer is a multiple of 3. Out of (n-1), n and (n+1), one has to be divisible by 3. If n is not divisible by 3, one of (n-1) and (n+1) must be divisible by 3.
Hence, (n-1)*(n+1) must be divisible by 3.

Using both statements, (n-1)*(n+1) must be divisible by 8*3 = 24. Remainder must be 0.

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

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New post 26 Sep 2012, 13:44
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ksharma12 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.



(1) n must be odd. Try n = 1 and n = 3.
Not sufficient.

(2) Try n = 1 and n = 2.
Not sufficient.

(1) and (2) together: When divided by 6, the remainders can be 0, 1, 2, 3, 4, 5. Since n is not divisible neither by 2, nor by 3, when divide by 6, n can give remainder 1 or 5, which means n is of the form 6k + 1 or 6k - 1, for some positive integer k.
If n = 6k + 1, then (n - 1)(n + 1) = 6k(6k + 2) = 12k(3k + 1) = 12k(k + 2k + 1). Since 2k + 1 is odd, k and 3k + 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n - 1)(n + 1) is divisible by 24.
If n = 6k - 1, then (n - 1)(n + 1) = (6k - 2)6k)= 12k(3k - 1) = 12k(k + 2k - 1). Since 2k - 1 is odd, k and 3k - 1 are of different parities, so their product is for sure divisible by 2. Altogether, (n - 1)(n + 1) is divisible by 24.
The remainder must be 0.
Sufficient.

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

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New post 12 Jan 2013, 03:34
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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

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

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New post 14 Sep 2017, 03:01
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By combining both the statement, we know that if a number is a prime number apart from 2, and 3, then it can be written in the form of (6n+1) or (6n-1) - plugging these values, we get that (n-1)(n+1) is always a multiple of 24; hence sufficient. Answer C.
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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1)  [#permalink]

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New post Updated on: 26 Mar 2013, 02:15
vbodduluri 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?
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.

Therefore answer is choice C.
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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1)  [#permalink]

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New post 31 Jan 2017, 17:12
1) n not divisible by 2=> n is odd=> (n-1) and (n+1) must be consective even numbers.

if n=1, 0*2/24 leaves remainder 0
if n=3, 2*4/24 leaves remainder 8
not sufficient


2) n not divisible by 3=> n can be even or 1, 5, 7, 11, 13....

if n=5, 4*6/24 leaves remainder 0
if n=2, 1*3/24 leaves remainder 3
not sufficient

together,
n must be odd and not divisible by 3=> n can be 1, 5, 7, 11, 13...
if n=7, 6*8/24 leaves remainder 0
if n=11, 10*12/24 leaves remainder 0

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

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New post 11 Sep 2017, 06:28
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What if n=1?

If so, (n-1) will = 0.

Please advise.

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

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

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New post 02 Jun 2018, 18:09
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


* n-1, n and n+1 are three consecutive integers, so one of them must be a multiple of 3, since every third integer is a multiple of 3. From Statement 2, n is not a multiple of 3, so one of n-1 or n+1 is.

* If, from Statement 1, n is odd, then n-1 and n+1 are even. Since every second even number is a multiple of 4, one of n-1 and n+1 is a multiple of 4 (at least) and the other a multiple of 2. So (n-1)(n+1) is a multiple of 8.

So with both Statements together we know that (n-1)(n+1) is a multiple of 3 and 8, and therefore of 24, and the remainder is zero.
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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1)  [#permalink]

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New post 15 Jun 2018, 08:30
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



Given , n > 0, r is remainder when (n-1)(n+1) is divided by 24, r = ?

(n-1)(n+1) is the product two consecutive even or odd integers, depending on whether n is odd or even.

Statement 1: n is not divisible by 2

n = odd, then we have, for n = 3, (n-1)(n+1) = (3-1)(3+1) = 8, hence r = 8
for n = 5, (n-1)(n+1) = (5-1)(5+1) = 4*6 = 24, hence r = 0

Statement 1 alone is Not Sufficient.


Statement: n is not divisible by 3.
hence, for n = 5, we have from above r = 0
& for n = 8, we have (n-1)(n+1) = (8-1)(8+1) = 7*9 = 63, hence r = 15

Statement 2 alone is Not Sufficient.

Combining the two statements, we get, n is not divisible by 2 or 3
for n = 7, (n-1)(n+1) = (7-1)(7+1) = 6*8 = 48, hence r = 0
for n = 11, (n-1)(n+1) = (11-1)(11+1) = 10*12 = 120, hence r = 0


Both Statements together are Sufficient.

Answer C.

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

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New post 25 Jun 2019, 09:57
Statement 1)
When n is not divisible by 2, then n can be 1,3,5,7,9
For n=1,3 & 5 , Different answers. Hence insufficient.

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

On combining these two statements, n is 1,5,7
For such numbers, the remainder is 0.

C is Correct
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