Aman verma wrote:Only few people attempted this question.
Harshavardhan said : it is never mentioned in the question that you cannot have 1 as the number. Now the number cannot possibly be 1 because if you divide 1 by any integer other that 1 and 0 you will always have 1 as the remainder.
Also when Harshavardha mentions even or odd ,he cannot be more wrong because the question has got nothing to do with even or odd. At best Harshavardhan's attempt at this question be called a conjecture which is not bassed on sound mathematical principles.
you've called for it.. ..here's my reply to few of the points that you've raised :
First, whenever you are posting a question on a public forum, asking others to respond, please make sure that you replicate the exact question, without adding your own "touch" . I'm sure that you've paraphrased the question and have not copy-pasted it because :
respective reminders,
for your information, it's always a remainder which is left when you perform division. "Reminder" is what you use when you don't trust your memory. Not your fault, it's the way you've been "trained".
Moreover the correct options IMO would say :
A) Either (i) or (ii)
B) Either (ii) or (iii)
.
.
A number cannot be n! AND n! -1 at the same time.
Now, as you have done these types of mistake at the first place, I cannot be more sure that there is some additional information in the question, which in your paraphrasing, has been omitted.
Now the number cannot possibly be 1 because if you divide 1 by any integer other that 1 and 0 you will always have 1 as the remainder.
Buddy, please read the question ( that you have written
) very very carefully.
It says, when you are dividing a number by 1, 2,3 .... n , it gives remainders as 0,1,... (n-1).
The thing to understand here is you are only dividing the number by 1 -> n .
Now when you have 1! as the number, n becomes 1 (n!) and hence, you have to divide 1! by 1 -> 1 ( i.e only by one number) and can you tell me what will be the remainder ? come on... you can do this calculation !!! remainder = 0 , which is (n-1).
Got it?! This satisfies the condition which YOU have mentioned.
which is not bassed on sound mathematical principles.
ha ha ha!! can't stop laughing !!! please.......
I'll quote another of your written sentences, and tell you a demo of sound mathematical principles :
because if you divide 1 by any integer other that 1 and 0
divide by zero?????????? what a sound mathematical principle!!!!!
now you too have joined me in the laughter riot!!!??? haven't you???
As far as I can see your explanation, I can only view it as a mugged-up explanation in the source from where you have tried to take this problem, reason being if you have genuinely tried to solve a mathematical problem ( mind you, it is different from practicing maths problem, which I am sure you have done in your school/engg) , you would have not said this :
the question has got nothing to do with even or odd
If you have a mathematical brain, you will know that there are multiple ways to solve a problem. Here as well, if you observe every value of 'n' will be odd. It cannot be even, reason being, the conditions put in the questions. Don't put stress on your brain, because it is said "it is better to avoid an activity where there is resistance".
In the end, I am not denying that the OA is B . It's just that I definitely think that some vital point, present in the original question, has been missed while replicating the question here.
Let's see if Mods/instructors can pitch in with their ideas.
