Remainder Prob

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by JS_2 » Mon Aug 29, 2011 10:05 pm
I just plugged in odd values in the expression :n^2-1 and concluded "A"

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by prashant misra » Thu Sep 01, 2011 10:22 pm
this simply an easy question i think.i did it by picking numbers evaluating both statements and got the right answer A

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by parul9 » Mon Sep 26, 2011 8:14 am
Answer is A.
What should be the ideal time to do this qsn?
I took abt 2 mins... is it a lot?
I think I could have done earlier, but I kept substituting values to be sure!

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by arunnarang » Sun Oct 09, 2011 5:09 am
here n^2 - 1 divided by 8.
it means when u put odd no like 1,3,5,7,9..it always give s u zero.
so answer is option 1.

but when u put even no like 2,4,6,8.its give u...remainder in the form of odd number.

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by parul9 » Sun Oct 09, 2011 5:28 am
IMO A

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by olylo » Mon Oct 17, 2011 5:10 am
The correct answer is A

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by immaculatesahai » Mon Nov 07, 2011 12:25 am
A wins.

You can factorize the expression to {(n-1)(n+1)}/8, this gives remainder of r.

Statement 1: n is odd. Meaning that n-1 is even and n+1 is even. Any 2 consecutive even numbers will have atleast 3, 2's in them. Hence remainder will be 0. You can pick arbitrary consecutive even numbers to check this.

Statement 2: Not Sufficient. You can try a few values of n which are not divisible by 8. Eg 12, 14 17 etc. Each will give different remainder.

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by SaeedaJ » Thu Nov 10, 2011 2:04 pm
I would say A.

1) When n is odd, the remainder is consistently zero. SUFF.
2) Plugging in even numbers for n results in various remainders. Inconsistent for a solution. INSUFF.

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by him1985 » Thu Jan 26, 2012 4:24 am
A is correct.
However i did it via conventional way...Thanx for different approaches
Himanshu Chauhan

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by preethikrishna » Thu Jan 26, 2012 10:54 pm
(A)

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by nisagl750 » Mon Feb 06, 2012 10:51 am
IMO its A

I checked it with n=3, 5, 7, 9 & 11

Though i forgot that if n is odd (n-1) & (n+1) would be two consecutive even integers always divisible by 8...So Stupid :cry:

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by hitesh.saini » Thu Feb 09, 2012 2:04 am
Hi Folks,

I am sorry but i would like to understand what this stands for! n^2-1

> is it n2

Because this symbol confuses me and when i look at yesterdays problem couldnt solve the same as well.

Any help in this regards would be appreciated.

/Hitesh.[/list]

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by MQ0451 » Sat Feb 11, 2012 1:31 pm
helps to have a large sample of squares memorized, but let's choose some at random. n = 1, the remainder is zero. n = 3, remainder is zero. n = 9, the remainder is zero. This seems to be the trend. Looks like we can conclude what the remainder will be (zero) when n is odd. Now if n is not divisible by 8, let's pick a couple choices. n = 3 we already solved, and we know the answer is zero. now if n = 2, the answer is three. We can't tell what the remainder is. Thus, choice 1 and only choice 1 helps us find the remainder - the answer is A.

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by neophytehemant » Tue Mar 06, 2012 3:00 pm
Hello,
Most of people said they concluded after checking few values, I choose option A when convinced that either (n-1) or (n+1)is multiple of 4 when n is odd and greater than 1.
Experts please suggest whether should we pick the answer choice or not after checking few values even if we are not 100% sure.

Regards,
Hemant

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by ka_t_rin » Wed Mar 14, 2012 11:56 pm
Absolutely agree with you) A) I`ve used the same logic)
augusto wrote:IMHO the answer is A.

1) Just plug in odd numbers,
n=1 : (1^2 - 1)/8 = 0
n=3 : (3^2 - 1)/8 = 0
n=5 : (5^2 - 1)/8 = 0
n=7 : (7^2 - 1)/8 = 0
after this I assume that r is always 0, so it�s sufficient

2) Again plug some numbers
n=1 : (1^2 - 1)/8 = 0
n=2 : (2^2 - 1)/8 = 3
upz, not sufficient.