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

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Bunuel
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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1) [#permalink] New post   03 Jan 2024, 00:50
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TargetMBA007 wrote:
Bunuel 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?

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.


Sorry, but I don't understand what yo mean. Please elaborate.
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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1) [#permalink] New post   03 Jan 2024, 04:28
Bunuel wrote:
TargetMBA007 wrote:
Bunuel wrote:
(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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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1) [#permalink] New post   03 Jan 2024, 05:45
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TargetMBA007 wrote:
Bunuel wrote:
(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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Re: If n is a positive integer and r is the remainder when (n - 1)(n + 1) [#permalink] New post   15 Jan 2024, 04:03
Bunuel wrote:
TargetMBA007 wrote:
Bunuel wrote:
(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.


Thanks as always Bunuel for all your help.
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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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