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

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

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03 Jul 2010, 22:19
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hI Everyone

Can someone, please explain to me, the following problem:

PROBLEM 2:
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.

Thanks for your help. I hope someone can explain to me.

regards
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27 Jun 2010, 21:26
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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.

Hope it helps.
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If n is a positive integer and r is the remainder when (n-1) [#permalink]

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27 Jun 2010, 20:31
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

OA:
[Reveal] Spoiler:
C
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Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]

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03 Jul 2010, 22:54
Question 2:

To be divisible by 24, a number must be divisible by 8 and by 3.

Statement 1: n is not divisible by 2. This basically says that n is an odd number. So if n is an odd number, both n-1 and n+1 are even numbers, which means that (n-1)(n+1) must be divisible by 4.

n-1 = 2k
n+1 = 2p

Between any two consecutive even integers, one has to be divisible by 4, so we know that the given number is divisible by 8.

Statement 2: n is not divisible by 3. However, between any three consecutive numbers one has to be divisible by 3. Considering the consecutive numbers (n-1)(n)(n+1) we know that n isn't divisible by 3, so either (n-1) or (n+1) should be divisible by 3.

Combining both together, we know that (n-1)(n+1) is divisible by 8 and 3 and hence 24. So reminder = 0. Hence answer is C.
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Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]

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04 Jul 2010, 05:21
in question 2, statement 2 alone will not be sufficient unless taken in conjunction with stmt 1 because if we consider ONLY stmt 2 - then 10 is a value which is NOT divisible by 3 and (10-1)*(10+1) = gives 99 which does not leave a remainder 0 when divided by 24. Thus IMO the answer to problem 2 should be A - stmt 1 alone is sufficient... pls correct me if I am wrong.
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06 Jul 2010, 12:45
@bunuel... 3 divided by 24 yields remainder of 3???
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06 Jul 2010, 17:34
bunnel:- i am too confused....how come 3 divided by 24 the remainder is 3?
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Re: NEED HELP - 2 DATA SUFFICIENCY PROBLEMS [#permalink]

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06 Jul 2010, 23:58
whiplash2411 wrote:
Question 2:

To be divisible by 24, a number must be divisible by 8 and by 3.

Statement 1: n is not divisible by 2. This basically says that n is an odd number. So if n is an odd number, both n-1 and n+1 are even numbers, which means that (n-1)(n+1) must be divisible by 4.

n-1 = 2k
n+1 = 2p

Between any two consecutive even integers, one has to be divisible by 4, so we know that the given number is divisible by 8.

Statement 2: n is not divisible by 3. However, between any three consecutive numbers one has to be divisible by 3. Considering the consecutive numbers (n-1)(n)(n+1) we know that n isn't divisible by 3, so either (n-1) or (n+1) should be divisible by 3.

Combining both together, we know that (n-1)(n+1) is divisible by 8 and 3 and hence 24. So reminder = 0. Hence answer is C.

n is not divisble by 2 implies n-1 and n+1 are even , this implies the product of these two numbers i divisible by 4 and not 8.
both the statements taken together say that the number is not even and not divisible by 3 , meaning all odd numbers not divisible by 3 .
these are 1,5,7,11,13,17--- , if u (square of any of these minus 1 )/24 = 1 and not 0
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07 Jul 2010, 03:46
utin wrote:
@bunuel... 3 divided by 24 yields remainder of 3???

ruchichitral wrote:
bunnel:- i am too confused....how come 3 divided by 24 the remainder is 3?

THEORY:
Positive integer $$a$$ divided by positive integer $$d$$ yields a reminder of $$r$$ can always be expressed as $$a=qd+r$$, where $$q$$ is called a quotient and $$r$$ is called a remainder, note here that $$0\leq{r}<d$$ (remainder is non-negative integer and always less than divisor).

So when divisor (24 in our case) is more than dividend (3 in our case) then the reminder equals to the dividend:

3 divided by 24 yields a reminder of 3 --> $$3=0*24+3$$;

or:

5 divided by 6 yields a reminder of 5 --> $$5=0*6+5$$.

Hope it's clear.
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Re: If n is a positive integer and r is the remainder when (n-1) [#permalink]

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01 Dec 2014, 14:09
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