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# If n is a positive integer and r is the remainder when

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

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24 Aug 2011, 00:55
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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) 2 is not a factor of n
(2) 3 is not a factor of n

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-n-is-a-positive-integer-and-r-is-the-remainder-when-n-98529.html
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Re: A complicated math prob. [#permalink]

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24 Aug 2011, 01:28
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1) n = 1,3,5,7 etc

n=7, 8*6 = 48, r = 0
n=9, 8*10 = 80, r = 8

NS

2) n=1,2,4,5 etc

n = 7, r=0
n = 10, r=3

NS

1+2) n=1,5,7,11, ..., 19

r = 0 in all the cases I tried, so I will go with C

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Last edited by pike on 24 Aug 2011, 02:20, edited 1 time in total.
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Re: A complicated math prob. [#permalink]

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24 Aug 2011, 01:32
statement (1) says 2 is not a factor of n. let n = 3, (n-1)(n+1) = 2*4/24 = 8/24 (remainder 8). let n be 5. then (n-1)n+1)/24 = 4* 6/24 (remainder zero). different remainders, insuffiecient.

statement 2 also insuffiecient, plug in value of n as 2 and 5 and we get different remainders.

combining both statements we have to take n as integer which is not a factor of 2 and 3. let us take n= 5. then (5-1)(5 +1)/24 remainder zero. again take n 7, then (7-1)(7+1)/24 = 8*6/24. remainder zero. so combining statement1 and 2 is sufficient to let us know that remainder of (n-1)(n+1)/24 is always going to be zero when n is not a factor of 2 or 3.

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Re: A complicated math prob. [#permalink]

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25 Aug 2011, 04:29
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Jasonammex 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) 2 is not a factor of n
(2) 3 is not a factor of n

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.

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

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20 Sep 2013, 23:15
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Re: If n is a positive integer and r is the remainder when [#permalink]

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21 Sep 2013, 03:06
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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?

Number plugging 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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-n-is-a-positive-integer-and-r-is-the-remainder-when-n-98529.html
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Re: If n is a positive integer and r is the remainder when   [#permalink] 21 Sep 2013, 03:06
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# If n is a positive integer and r is the remainder when

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