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

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.

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

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 = 3⁴  = 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.

Again Some thing newWhy

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.

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.

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

(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.

misconception ! 2! + 3! is

misconception !

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