Problem Solving

Aman verma wrote: Moreover Harsha don't have to worry about my brain as their is neither any resistance nor any stress, he better be bothering about his own highly appreciated mathematical brain.

is this a sentence correction test for us? If yes, there is another forum for it!

"Harsha" is singular, and will require doesn't .

their is neither any resistance nor any stress

Their is used to show possession. You should use There instead.

Now I'm 100% confident that it was not a typo.

Regarding the 'reminder ' part of the question it was only a typing error.

there are a looooooot more... but, as the source is not reputable, my efforts would be futile in its correction. Plus this is a Math forum.

Mods, I apologize for this argument, and will stop now. But when the object is requesting it, the subject has to do its verb.

I think moderators should take care of this. And having talked about grammatical errors we are not here to show our grammatical skills and typing errors could always occur.Now harsha can rest assure when he will recieve an abuse it will be free from any grammatical error or typing error.And he will receive the abuse free from grammatical error in this forum itself ,no need to go anywhere else .When the subject is so keen the object will do its final verb.
Ha ha ha

Aman verma wrote:I think moderators should take care of this. And having talked about grammatical errors we are not here to show our grammatical skills and typing errors could always occur. When the subject is so keen the object will do its final verb.
Ha ha ha

I will just say that you did not adhere to the rules set by the moderator:

https://www.beatthegmat.com/post72205.html#72205

Look who's talking ! Let the Moderators be the better judge.

Here in the problem we find when the no. is divided by n it leaves remainder (n-1).
so we can write number of the form kn+(n-1) = (k+1)n-1which when divisible by n gives a remainder of (n-1). where k=1,2,3....
let m=k+1, so the required no. is of the form mn-1 where m=2,3,4........

now if we compare the given options
(i) is n! which is not of the form mn-1
(ii) is n!-1 which seems of the form mn-1
so we write n!-1 = n(n-1)!-1,
so for n!-1 to be the solution {(n-1)!=m}>or =2, which is true only for values n>3 , it is not applicable for n=1,2.

(iii) LCM of (1,2,3,....n)-1 = (1.2.3.4......n)/(no. which is common further) - 1, but it is also not applicable for n=1,2 as above..its similar.


so overall we can say the solution can be determined and it is mn-1 where m= integers>1.
but none of the given ans. is the solution.

Hope I have solved the problem ..

Here the answer is B

If x (for example 15) is divisible by a (for example 3), then when (x-1) (14 in our example) is divided by a the remainder is (a-1) (it is right, 2).

In the problem above, when possible number is divided by any of the 1,2,...,n sequence the remainder is 1 less that the divisor itself (when divided by 1 remainder is 0, by 2 - remainder is 1, etc)

We know that n! is divisible by any number in the sequence of 1,2,...,n. So, if (n!-1) is divided by any number between 1 and n the remainder is 1 less than that number. (ii) can be the number we need.

The same is applicable to (iii), because LCM (1,2,...,n) is divisible by any number between 1 and n. So, (iii) can be the number we need.

n! cannot be the number we need, because when we divide it by the numbers between 1 and n the remainder is always 0.

The correct answer is B.

If you have internalized the concept and logic of factorials, LCM, then you can solve this problem orally!

Shashank.ism confused n which he defined at the beginning of his solution with n defined in the answer!

If we write down his nm-1 formula as jm-1 we can see that (ii) and (iii) are possible. For any number between 1 and n denoted j n! can be written as j * n!/j where n!/j is integer, because n! is the product of 1,2,...,n set, which already includes j. If we replace n!/j with m we arrive at the formula n!-1 = jm - 1. The same method is applicable to (iii).

Eldorjon wrote: n! cannot be the number we need, because when we divide it by the numbers between 1 and n the remainder is always 0.

Eldorjon, I agree with you for most of the explanation, but my question was : why can't we view 1 as 1! and then divide it from 1->1 (1->n) i.e. by a single number. What we get as remainder is 0, which is n-1, and which satisfies the condition.
This is one boundary condition, which should not be missed in questions like these.

Please correct me if I'm wrong.

Eldorjon wrote:Shashank.ism confused n which he defined at the beginning of his solution with n defined in the answer!

If we write down his nm-1 formula as jm-1 we can see that (ii) and (iii) are possible. For any number between 1 and n denoted j n! can be written as j * n!/j where n!/j is integer, because n! is the product of 1,2,...,n set, which already includes j. If we replace n!/j with m we arrive at the formula n!-1 = jm - 1. The same method is applicable to (iii).

Eldorjon I am talking about mn+1 where m has already got a condition that m>1
since I have replace K+1 by m where k=1,2,3,4...........
now hether u replace m by j ,p,q anything it doesn't matter the condition will accordingly apply ie. j>1, p>1, q>1....with whichever u replace.
but if we take all the values of k i.e. -ve integers too...then we can say that m can have any values (including _ve integers..)

In that case obviously (B) would be the answer..