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If n is a positive integer and r is the remainder when n^2 - 1 is

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If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 21 Jan 2012, 18:04
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If n is a positive integer and r is the remainder when n^2 - 1 is divided by 8, what is the value of r?

(1) n is odd
(2) n is not divisible by 8



As the OA is not provided this is how I solved this. Please let me know whether I am right or not.

Considering Statement 1

n =1 then remainder will be zero. ----------------> Is my thinking correct over here? If it is then for me this statement is sufficient to answer the question.

Considering statement 2

There will be different values for r and therefore its insufficient.

Therefore, for me the answer should be A unless someone has any other ideas. Please help.

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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 21 Jan 2012, 18:25
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If n is a positive integer and r is the remainder when n^2 - 1 is divided by 8, what is the value of r?

n^2-1=(n-1)(n+1)

(1) n is odd --> both n-1 and n+1 are even. Moreover, they are consecutive even integers thus one of them is divisible by 4 too. Now, as one is divisible by 2 and another by 4 then (n-1)(n+1) is divisible by 2*4=8. Sufficient.

(2) n is not divisible by 8 --> try n=1 to get an YES answer and n=2 to get a NO answer. Not sufficient.

Answer: A.
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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 21 Jan 2012, 18:45
enigma123 wrote:
If n is a positive integer and r is the remainder when n^2 - 1 is divided by 8, what is the value of r?
1). n is odd
2). n is not divisible by 8

As the OA is not provided this is how I solved this. Please let me know whether I am right or not.

Considering Statement 1

n =1 then remainder will be zero. ----------------> Is my thinking correct over here? If it is then for me this statement is sufficient to answer the question.

Considering statement 2

There will be different values for r and therefore its insufficient.

Therefore, for me the answer should be A unless someone has any other ideas. Please help.


As you can see from my post above answer is indeed A. But you arrived to A while trying only one value for n, which is not enough. For some other question, different values could give different answers and a statement would be insufficient in that case. What I mean is, when you decide that a statement is sufficient based only on plug-in method you should make sure that you tried several different numbers (and saw some pattern maybe), and even in this case you may not be 100% sure that the answer would be correct. Though if several numbers give the same answer and you are able to see some pattern, then you can make an educated guess that a statement is sufficient and move-on.

Generally on DS questions when plugging numbers, your goal is to prove that the statement is not sufficient. So you should try to get an YES answer with one chosen number(s) and a NO with another.

Hope it's clear.
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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 24 Jan 2012, 18:07
Bunuel - its clear apart from one minor doubt. In statement 1 why can't we take n =1 ?
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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 24 Jan 2012, 18:19
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enigma123 wrote:
Bunuel - its clear apart from one minor doubt. In statement 1 why can't we take n =1 ?


Actually we can. Solution above does not exclude this possibility: n=1=odd then n^2-1=0 and zero is divisible by any integer (except zero itself), so it's divisible by 8 too. Sufficient.

The point is that you arrived that (1) is sufficient based only on one value of n, n=1. And as I discussed above one value is not enough to conclude that the statement is sufficient.

Hope it helps.
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New post 24 Jan 2012, 18:24
Oh yes. Entirely agree. Many thanks again.
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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 26 Dec 2016, 07:49
1). N is odd, then n=2k+1, n^2 - 1=(2k+1)^2-1=4k^2+4k=4k(k+1). One of k and k+1 must be even, therefore, 4k(k+1) is divisible by 8.
Answer is A
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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 26 Jun 2018, 11:25
If n is a positive integer and r is the remainder when n2 - 1 is divided by 8, what is the value of r?
1). n is odd 2). n is not divisible by 8

n-1,n,n+1
(n-1)(n+1) is divided by 8 so (n-1)(n+1) is even ,hence n is odd
1). n is odd suff.
2). n is not divisible by 8 extra information :-o
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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 29 Mar 2019, 06:56
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enigma123 wrote:
If n is a positive integer and r is the remainder when n^2 - 1 is divided by 8, what is the value of r?

(1) n is odd
(2) n is not divisible by 8


Given: r is the remainder when (n² - 1) is divided by 8

Target question: What is the value of r?

Statement 1: n is odd
Let's test some ODD values of n
If n = 1, then n² - 1 = 1² - 1 = 0, and 0 divided by 8 leaves remainder 0. So, the answer to the target question is r = 0
If n = 3, then n² - 1 = 3² - 1 = 8, and 8 divided by 8 leaves remainder 0. So, the answer to the target question is r = 0
If n = 5, then n² - 1 = 5² - 1 = 24, and 24 divided by 8 leaves remainder 0. So, the answer to the target question is r = 0
If n = 7, then n² - 1 = 7² - 1 = 48, and 0 divided by 8 leaves remainder 0. So, the answer to the target question is r = 0
At this point, we might conclude that r will ALWAYS be 0
So, statement 1 is SUFFICIENT

----ASIDE--------------------------------
If you're not convinced, here's an algebraic solution as well:
If n is ODD, then n = 2k + 1 (for some integer value of k)
So, n² - 1 = (2k + 1)² - 1 = 4k² + 4k + 1 - 1 = 4k² + 4k = 4(k² + k)

Notice that, if k is odd, then k² + k is EVEN, which means k² + k = 2 times some integer
So, n² - 1 = 4(k² + k) = 4(2 times some integer) = 8 times some integer
In other words, n² - 1 is a multiple of 8, which means the answer to the target question is r = 0

Similarly, if k is even, then k² + k is EVEN, which means k² + k = 2 times some integer
So, n² - 1 = 4(k² + k) = 4(2 times some integer) = 8 times some integer
In other words, n² - 1 is a multiple of 8, which means the answer to the target question is r = 0

In both cases, the answer to the target question is r = 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
------------------------------------------

Statement 2: n is not divisible by 8
There are several values of n that satisfy statement 2. Here are two:
Case a: n = 3. In this case, n² - 1 = 3² - 1 = 8, and 8 divided by 8 leaves remainder 0. So, the answer to the target question is r = 0
Case b: n = 4. In this case, n² - 1 = 4² - 1 = 15, and 15 divided by 8 leaves remainder 7. So, the answer to the target question is r = 7
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 30 Mar 2019, 09:29
#1
check for n=1,3,5,7 ; r= 0 always
sufficient
#2
n=odd + even integers 2,4,6..
insuffciient
IMO A

enigma123 wrote:
If n is a positive integer and r is the remainder when n^2 - 1 is divided by 8, what is the value of r?

(1) n is odd
(2) n is not divisible by 8



As the OA is not provided this is how I solved this. Please let me know whether I am right or not.

Considering Statement 1

n =1 then remainder will be zero. ----------------> Is my thinking correct over here? If it is then for me this statement is sufficient to answer the question.

Considering statement 2

There will be different values for r and therefore its insufficient.

Therefore, for me the answer should be A unless someone has any other ideas. Please help.

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Re: If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 30 Mar 2019, 09:58
In other words,

{(Any odd integer)^2-1} is always DIVISIBLE by 8.

Because:
Odd number= (2k+1)
Odd number^2=(2k+1)^2=4k^2+4k+1
Thus, Odd number^2-1= (4k^2+4k+1)-1=4k^2+4k=4k (k-1)
Here, since k is ODD, K-1 MUST BE EVEN & 4K is DIVISIBLE BY 4.
Thus, full expression is definitely divisible by 8.

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If n is a positive integer and r is the remainder when n^2 - 1 is  [#permalink]

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New post 30 Mar 2019, 10:16
In other words,

{(Any odd integer)^2-1} is always DIVISIBLE by 8.

This can also be proven in following way:
Assume x= a odd number
x^1-1= (x+1) (x+1)
Here, since x is an odd number
(m-1) is an even and (m+1) is another even number. More importantly, (n-1) and (n+1) are two consecutive even numbers.
For example, n=3,
So, n-1=2 & n+1= 2.

Now come to the theory, product of two consecutive even number is always divisible by 8 because one of these two consecutive even numbers is divisible by 2 and the other number must be divisible by 4.

Thus, (x^2-1) is always divisible by 8 if x is an odd number

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If n is a positive integer and r is the remainder when n^2 - 1 is   [#permalink] 30 Mar 2019, 10:16
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