# NO. SYS QN2??

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 !!!

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 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 sol. However in this particular problem we need not go throug such a hassle.

3^{450} = x mod(3^{3}. 4)

=> 3^{447} = x/3^{3 }mod(4)

it can be easily shown from periodicity of power residues that x/3^{3 }= 3 => x = 3^{4 }= 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.

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

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 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.

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

Thanks for the above.

misconception !

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

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