NO. SYS QN2??

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neel_saxena's picture
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QA) WATS THE REMAINDER OF (100!+1)/101?.....ANS-0
QB) WATS THE REMAINDER OF (3^450)/108??....ANS-81
QC) Mr. X added "N" no's & got the sum as 1850.However he added a no. twice by mistake.what is the difference b/w N & that no.??...ans-40

HELP ME !!!

anita1's picture
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QB. Following Rajorshi's

QB. Following Rajorshi's method in post http://cat4mba.com/node/2515#comment-929

108 = 2 x 54 = 4 x 27 (4 and 27 are co-prime)

N = 1 (mod 4)

N = 0 (mod 27)

Now we need to find a number U s.t  27U = 1 (mod 4)

                                                                   => U =3

Similarly for 4V = 1 (mod  27) we get V = 7

Thus r= 1 x 27 x 3 + 0 x 4 x  7 = 81

Thus N = 81 (mod 108)

Or the reminder is 81

 

cat2007's picture
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ans to ur ques

hi!

Q1) For any prime number "a"

[(a-1)! + 1] is always divisible by a
so for prime number 101 , (101-1)! + 1 = 100! + 1 is divisible by 101, so, there is no rem.
therefore ans is 0.

secondharsa's picture
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I think the question should

I think the question should be

Mr. X added first  "N" no's & got the sum as 1850 . . .

Sum of first 60 Numbers is 1830

and Mr.X added 20 twice to get it as 1850

required answer = 60 - 20 = 40

rajorshi's picture
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Anita was spot on with her

Anita was spot on with her sol. However in this particular problem we need not go throug such a hassle.

3450 = x mod(33. 4)

=> 3447 = x/33 mod(4)

it can be easily shown from periodicity of power residues  that  x/33 = 3 => x = 3= 81

the method that anita used was used by me for a problem where there was no way i could simplify the problem by taking out common factors. This time its way easier.

 

 

anita1's picture
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Again Some thing newWhy
Again Some thing new
Why IMS/CF don’t provide all these tricks/tips in their material.

I have the BRM and materials of both and no where they have used the mod concept  and many other stuffs that I learnt from you guys

 

 

rajorshi's picture
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i guess the whle point is to

i guess the whle point is to do some reading on one's own .  spoon feeding just would not be enough.

Sid87 (not verified)
Sid87's picture
But at least you should know

But at least you should know what to where and where to read. I have completed my english honours and now preparing for CAT. After attending all the classes of IMS I feel they lack the quality in providing the study notes. I never face any problem in understanding the concepts given in BRM and questions given after each class. But the questions gven in set B are in differnt zone altogether. No detail explanation , no notes, just providing solutions never help in understanding the concepts. I really feel frustrated and now after attempting FLT 7 on this site things are getting worse.

singh_yash's picture
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another solution 2 part b

Its not imperative to know 'mod. method' to solve the second part,if one knows its well and good ;but if one doesn't, familiar methodologies can be used to solve it !!

3^450/108 =3^450 / (3^3)(2^2) =3 ^ 447 / 4 =3 (3^446) / 4 =3( 9^223) / 4 = 3(1+8)^223/ 4 

now each term containing 8 would be divisible by four hence  the only term which remains is 3((1)^223)=3

but we have cancelled out 3^3 from the actual fraction hence the actual remainder is (3^3)x3=81

A simple example to understand the above concept is as follows:

what is the remainder when 64 is divided by 6? , we all know it is 4.

mathematically , 64/6=32/3 ,32/3 gives remainder=2, i.e. to get the correct remainder we have to multiply (the obtained remainder) by 2 as 2 was cancelled from the actual numerator and denominator. 

For the first part, (a!+b!) is divisible by a+b ,hence (100!+1!)=(100!+1) is divisible by (100+1)=101, implies remainder is zero

secondharsa's picture
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(a!+b!) is divisible by a+b

(a!+b!) is divisible by a+b ,hence (100!+1!)=(100!+1) is divisible by (100+1)=101, implies remainder is zero

Thanks for the above.

rajorshi's picture
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misconception ! 2! + 3! is

misconception !

2! + 3! is not divisible by 5 , there are million other examples.

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