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Remainder Prob
this simply an easy question i think.i did it by picking numbers evaluating both statements and got the right answer A
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!
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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arunnarang
- Newbie | Next Rank: 10 Posts
- Posts: 3
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.
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.
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.
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.
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.
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.
A is correct.
However i did it via conventional way...Thanx for different approaches
However i did it via conventional way...Thanx for different approaches
Himanshu Chauhan
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hitesh.saini
- Newbie | Next Rank: 10 Posts
- Posts: 1
- Location: Stockholm, Sweden
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]
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]
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.
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
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
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.
