thanks all you ppl for all the answers..but last ques..remainder is 7!
A:6 I also think that remainder should be 1
(1+13*n)/13 is the last exp, which implies that remainder shud be 1
"2(8!)-21(6!) divides 14(7!)+14(13)!?"
Is the factorial out side of bracket ?
nimisha_2009 Saysthanks all you ppl for all the answers..but last ques..remainder is 7!
arre yaar can u post the soln for this....
its cming 1......
???
nimisha_2009 Saysthanks all you ppl for all the answers..but last ques..remainder is 7!
Q6. what is the remainder when 2(8!)-21(6!) divides 14(7!)+14(13)!?
2(8!)-21(6!)= 6!(112-21)=6!(91)
%=0
%= %= 7!
Hence remainder is 7!
Hi can anyone send me the link for Arun Sharma ebook? please its very urgent, PM me in case u have.
swami_84 SaysHi can anyone send me the link for Arun Sharma ebook? please its very urgent, PM me in case u have.
Hi swami!
I will suggest you to buy the book. In my personal opinion the book is worth it. 😃 It will cost you around 200 bucks!
Otherwise, you can try borrowing it from your mates, or find the second hand book stalls. And keep discussing the sums on this thread itself!
I guess, finding the electronic version is a difficult task!
hi friends.i want to know which book will be the best for solving DI n LR?
i was previously solving from my coaching stuff which had very few problems and of fewer types.
i heard a book of DI n LI by arun sharma.i want to know if buying that book is worthwhile and will it be sufficient.
if not,then guys please suggest me some good way.
Som books for Di/lr
1.Arun sharma DI/LR
The focus is more on DI rathr than LR
iF u really wannna improve DI skills ,Its the Perfect book
2. Nishit sinha Di/LR(pearson publicatiobns)
Heard a lot abt the book but didnt get ny hands on xpo
so bttr u chk it urself
3.I recommend AImcats(TIME mock cat)
I promise u will face the toughest LR problems thr
try 2 collect last yr aimcat papers 2 get max. xpo
So conclusion DI-arun sharma
LR-aimcats
hi friends.i want to know which book will be the best for solving DI n LR?
i was previously solving from my coaching stuff which had very few problems and of fewer types.
i heard a book of DI n LI by arun sharma.i want to know if buying that book is worthwhile and will it be sufficient.
if not,then guys please suggest me some good way.
i will suggest u to do 1 thing...do mock test's DI LR and i have a file for DI qs..if u want i can mail u....and solve DI and LR questions over pg : i think this is enough :
nidhisoni24
am havin tough time clearing the DI part, though am a bit better in LR.
Can plz u send me the DI ques([email protected])....:smile:....am also not so good in quants , mostly num system, quad eqn , time speed dist....can u suugest something for this.....
hi friends.i want to know which book will be the best for solving DI n LR?
....
if not,then guys please suggest me some good way.
Hey loop,
First thing first, Mind it, you should have shoot your query in some other relevant or near-to-relevant thread(s). 😃 Anyways.. !!
Again First thing First, did you try getting some idea on the type of DI sets which have been asked in the recent past. If you haven't done that, I will suggest you to do that ASAP and let you learn yourself, where and how you need to work. As far as the specific book is concerned, I don't have any list, but practicing from wherever and whatever you can, will be the best option. Arun Sharma's book is good as well. But timed attempts on DI sets, as much as you can, will be of great help, trust me! 😃 Pick up, previous year papers, mock papers and some work-sheets. Start solving! :)
Regards,
maninblue :thumbsup:
P.S. I am not a good scorer in this particular section, :sad:, but I am a good counselor!
nidhisoni24
Hi..nidhi can u mail me the di sets too to my mail id....
My mail id is [email protected] k....
Hi,
I have a doubt in one of the concepts given in the book in the chapter Permutations and Combinations.
-If n identical things are to be distributed among r persons, such that each may get any number of things. This can be done in n-r-1 C r-1 ways .
I will appreciate if anyone can explain this concept.
It is actually n+r-1 c r-1 ... suppose you have to distribute 7 apples among a,b, c, d four people. that means a+b+c+d=7.. imagine this situation as arranging 7 zeros and 3 ones... like (0 0 0 0 0 0 0 )to distribute this 7 zeros in to 4 groups you need three separators(say 3 ones). one arrangement can be 0 1 0 1 0 0 0 0 0 ....So the question reduced to arrangement of seven zeros and 3 ones in a line... so it is 10!/(7! * 3!) so this is 10c3 it is( n+r-1) C (r-1)
smrutirekha SaysIt is actually n+r-1 c r-1 ... suppose you have to distribute 7 apples among a,b, c, d four people. that means a+b+c+d=7.. imagine this situation as arranging 7 zeros and 3 ones... like (0 0 0 0 0 0 0 )to distribute this 7 zeros in to 4 groups you need three separators(say 3 ones). one arrangement can be 0 1 0 1 0 0 0 0 0 ....So the question reduced to arrangement of seven zeros and 3 ones in a line... so it is 10!/(7! * 3!) so this is 10c3 it is( n+r-1) C (r-1)
Kool Explanation... 7 zeros and 3 ones each as a separator! Thanks a lot...
Since u people discussing about P&C;, I have a query.
How v can find the no. of quadrilaterals from x no. of points out of which y are collinear?
I don't remember the question no properly, but i guess its from LOD1 of the same chapter.
Since u people discussing about P&C;, I have a query.
How v can find the no. of quadrilaterals from x no. of points out of which y are collinear?
I don't remember the question no properly, but i guess its from LOD1 of the same chapter.
@vishrock
The question you are talking about is like there are 25 points out of which 7 are co-linear. How many quadrilaterals can be formed using these 25 points.
See to form a quadrilateral we need to select 4 points from these 25 points.
This can be done in (25 C 4 )ways.
We cannot form a quadrilateral if more than 2 points that are selected is co-linear. So we will subtract those cases when 3 or 4 of the points are co linear.
3 collinear points can be selected in (7 C 3) * ( 25-7 C 1) ways
4 collinear points can be selected in (7 C 4) ways
So final answer will be (25 C 4 ) - (7 C 3) * ( 25-7 C 1) - (7 C 4)
Since the answer to this problem is none of the above. So cant verify my answer but still I think this solution should be fine.
what is the highest power of 3 available in the expression 58!-38!
a) 17
b)18
c)19
d)none of these
what is the highest power of 3 available in the expression 58!-38!
a) 17
b)18
c)19
d)none of these
58/3=19
19/3=6
6/3=2
38/3=12
12/3=4
4/3=1
So 58!-38! can be written as 3^27*x-3^17*y where x and y are not divisible by 3
3^17(3^10*x - y)
Now how do i go about evaluating if the term in the bracket will have a factor of 3?:huh:
58/3=19
19/3=6
6/3=2
38/3=12
12/3=4
4/3=1
So 58!-38! can be written as 3^27*x-3^17*y where x and y are not divisible by 3
3^17(3^10*x - y)
Now how do i go about evaluating if the term in the bracket will have a factor of 3?:huh:
Hey i think the answer would be 3^17 only as u have extracted the powers of 3 so x and y can not be 3 or a multiple of 3
so in that case they wont be divisible by 3
hence 3^17
I would also like to ask divyaakanksha to post the solution or answer to the above problem in case she has it other wise we can ask quant devils like implex or made_for_iims people who can confirm the asnwers but i think it is fine
lets c what others have to say