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655-705 Level|   Remainders|      
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Bunuel
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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 03 Jan 2024, 00:50
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Bunuel

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?

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.

Hi Bunuel, With statement 1, if we assume n = 1, then, why can we still assume that n is a multiple of 8, when we solve together? My guess is, we cant, but if we were to assume n = 1, we still get r = 0 (which is the same case as when we assume n is a multiple of 8 and 3), and hence, we still consider this sufficient together? Would just like to validate my thinking on this.

Bunuel.
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Sorry, but I don't understand what yo mean. Please elaborate.
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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 03 Jan 2024, 04:28
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Bunuel

(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), ...).
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Sure, you mention here that "n must be divisible by 8", i.e. n = 8x.
But in this statement, if we assume n = 1, then (n-1) (n+1) would be 0*2=0. So, I wonder in this case, why can we still assume that n is a multiple of 8, when we solve the two statements together?

My guess is this is because, even if n = 1, we still get r = 0 (which is the same case as when we assume n is a multiple of 8 and 3), and hence, the statements would still be sufficient together? Would just like to validate my thinking on this.

Thanks
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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 03 Jan 2024, 05:45
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Bunuel

(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), ...).

Sure, you mention here that "n must be divisible by 8", i.e. n = 8x.
But in this statement, if we assume n = 1, then (n-1) (n+1) would be 0*2=0. So, I wonder in this case, why can we still assume that n is a multiple of 8, when we solve the two statements together?

My guess is this is because, even if n = 1, we still get r = 0 (which is the same case as when we assume n is a multiple of 8 and 3), and hence, the statements would still be sufficient together? Would just like to validate my thinking on this.

Thanks
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First of all, the solution there says that \((n-1)(n+1)\) must be divisible by 8, not n. Next, if n = 1, then \((n-1)(n+1)\) becomes 0, which is divisible by every integer except 0 itself.
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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 15 Jan 2024, 04:03
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Bunuel
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Bunuel

(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), ...).

Sure, you mention here that "n must be divisible by 8", i.e. n = 8x.
But in this statement, if we assume n = 1, then (n-1) (n+1) would be 0*2=0. So, I wonder in this case, why can we still assume that n is a multiple of 8, when we solve the two statements together?

My guess is this is because, even if n = 1, we still get r = 0 (which is the same case as when we assume n is a multiple of 8 and 3), and hence, the statements would still be sufficient together? Would just like to validate my thinking on this.

Thanks

First of all, the solution there says that \((n-1)(n+1)\) must be divisible by 8, not n. Next, if n = 1, then \((n-1)(n+1)\) becomes 0, which is divisible by every integer except 0 itself.
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Thanks as always Bunuel for all your help.
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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 13 Feb 2025, 22:47
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how come 8 divided by 24 yields remainder of 8.
Bunuel
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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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 14 Feb 2025, 00:00
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how come 8 divided by 24 yields remainder of 8.
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Let me ask you a question: if you had 8 apple and wanted to distribute it evenly in 24 baskets, each basket would get 0 apples and 24 apples would be leftover (remainder).

When a divisor is larger than the dividend, then the remainder is the same as the dividend. For example:

3 divided by 4 yields a remainder of 3: 3 = 4*0 + 3.
9 divided by 14 yields a remainder of 9: 9 = 14*0 + 9.
1 divided by 9 yields a remainder of 1: 1 = 9*0 + 1.
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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 28 May 2025, 04:02
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how i approached it

(1) not divisible by 2, so odd
(2) not divisible by 3, so either even, 5, 7

1+2
its either multiple of 5 or 7
so now
n^2-1 = 24k + remainder
hence,
25-1 = 24 remainder = 0
625-1 = 624 , remainder = 0

for 7,
49 - 1 = 48, remainder is 0
49*49 -1 = ( 49*40 + 49*9 )- 1= 2401 -1 = remainder = 0

hence suff.
enigma123
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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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 13 Sep 2025, 02:27
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hi Bunuel,
here u said, insufficient on its own (highlighted part), it means?
Bunuel
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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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 13 Sep 2025, 02:50
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shaliny
hi Bunuel,
here u said, insufficient on its own (highlighted part), it means?

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“Insufficient on its own” means that the statement alone is not sufficient to answer the question. However, even though it’s not sufficient, we can still draw some useful conclusions from that statement, and those can be applied when we evaluate the two statements together.
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If n is a positive integer and r is the remainder when (n - 1)(n + 1) 26 Sep 2025, 02:12
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Hi Bunuel,

How to master question types like these, I am really struggling in the approach and not understanding what fundamentals are missing here? :(
Looking forward to your response or help from any expert in the community who can help resolve the root cause of issue here.

Best,
Prathamesh Patil

SΙC•PARVΙS•MAGNA
Greatness from Small Beginnings


Bunuel
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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