This extremely useful article for finding remainders has been contributed to TG by Fundoo Bond, another intelligent TGite. All the CAT 2007/2008 who find this article useful are requested to say thanks to Fundoo Bond for his efforts and this wonderful compilation. If you have some interesting article, funda or helpful information to share, please mail it to us and we will post it by your name- Total Gadha

The problems of finding the remainder are considered the most dreadful among the number theory problems. Fortunately, if we arm ourselves with some basic theorems, we will see that we can turn these mind-boggling problems into sitters and can ensure some easy marks.

I will be trying to share both the theory and the examples in order to make the concepts clearer. Again friends, do keep in mind that I am not a genius to write these theorems on my own. I have learnt them from various sources and I am trying my best to explain you in my own words.

Thanks Sir...This is what I was searching for...and you have given it in the best time...

You are really TG.....Toooooooo Gooooood

with these Fundoo theorams, you have proved yourself BOND

hey folks

how to do 40!%83

hi Mr. Fundoo Bond.

this is a masterpiece 2 learn the concepts of remainder theorem.i found the chinese remainder theorem a bit more abstruse 2 master,yet other theorems and their enunciation with examples proved really helpful 2 me.hope 2 see such great work again from u.

HATS OFF TO U

Hi TG,

really an unbelievable article to say the least ....

Continue your good work ...

Thanks,

Jitendra

hi fundoo,

Though we all know you are a fundoo and a bond.....by sharing this article to have shown altruistic character.Hats off to you.

Thanks

Amit

Hi Fundoo,

Wonderful article and great compilation!

Thanks!

S P

i'm sure u wont believe it ...but was just going to post a request to you asking you to explain these very same theorems ....had seen them in one of your post wherein tg mentioned these are your favourite topics:>

secondly wanted to thank you for all your previous posts...u really took grt pains in explaing all your methods in detail( remember the one in which i asked for digital root and you typed up the detailed approach) and now this....wht can i say but 'rock on dude'

cheers!!

Hi Fundoo,

Its really superb.Only one thing I can say "Fundoo".

Thanks and Regards,

Manish

hi fb,

First, thank you for the post. It is very helpful.

In the ques. :  how to solve 
 
 (ii) remainder of 2 ^1990 / 1990
1990=2.5.199
from Euler's Theorem
phi(2)=1
phi(5)=4
phi(199)=198
2^4=1 mod 5--------(1)
2^198=1 mod 199---------(2)
LCM(4,198)=396
so
2^396=1 mod 1990
2^1980=1 mod 1990(because 1980 is a multiple of396)
so now we remain with
2^10 mod 1990
1024 mod 1990=1024
ans is 1024

how can we apply Éuler's theorem in this case? because according to ur post we can only use the theorem when both M(=2) and N(=1990) are coprimes, but this not the case here.

Kindly elaborate.

hey,

thnx for this article...i hav been lukin for the totient style of solving method since last few months

the other methods were as fascinating...

......it was extremely helpful...

thnx

rohit

hey,

thnx for this article...i hav been lukin for the totient style of solving method since last few months......it was extremely helpful...

thnx

rohit

@AD P - Thanks for the prompt response.

Hi Aviator D
I followed the Euler's Theorem
Euler's Theorem is the generalization of Fermat's Little Theorem, applicable when ur dividend is composite in form.So we first find prime factors for the composite no.
-----After getting prime factors, use Euler's totient to obtain phi(x) values. phi(x)
just define no of no which are coprime to x & <x
-----After getting totient for each prime factor, apply Fermat's little or Euler for each prime factor
Here 1990 is composite
jo first convert it into primes(or prime factors)

Hi AD P,   

Refering to your post, i understand that we can use Euler's theorem when the dividend(N,here) is a composite no., but additonally is it not required that we ensure that divisor(M,here) is relatively coprime to N ?

finding out the totient for each prime factor of 1990(i.e. 2*5*199) would give us the remainders when 2^1990 is divided by each of 2^1,5^4 and 199^198, how do we then use these remainders obtained to get the final remainder when 2^1990 is divided by 1990 ?

Hi Fundooooooooo, just want to say

Tooooooooooooooooooooooooooo Goooooooooooooooooooooooooooooooooooooooooood

helllooo sir ..

thankss for  such a excellent concept but sir  i cldnt be able to take out the print of ths page !! wat shld i do now ?

Thank you,sir,the article is amazing.i was searching for these concepts from a long time...god bless u guys, u guys r doin a damn gud job...some fundas on algebra,would be of great help,especially on hw. to solve prob.on graphs and solving prob.with the help of graphs...thankin u .

harshal.

This is really a great stuff...and this site is cool

hanks a lot...

really a grt..stuff...

was actually looking for this one...in a compiled format...

and...AAAAHHH!!! here it is ..

thx fundoo..

i dnt mean to say that

41! mod 83 cant be -1

yes it  can be -1 or 1 

n thus i predicted ans is either 2 or 81

bt which one is correct i dnt knw

I have a big doubt on your approach. 2^4 mod 5=1 and 2^198 mod 199=1 doesn't mean that 2^396 mod 1990=1. It can even be checked by the fact that 2^396 and 1990 are both even and hence the remainder should be even. My solution is like this -

2^1990 mod 1990 = 2^1989/(199*5)

using chinese remainder theorum,

2^1989 mod 199(A) = 114(r1)

2^1989 mod 5(B)= 2(r2)

for 199x+5y=1, x=-1 and y=40

hence remainder = 199*2*-1+114*5*40=22402

or, 22402-995*22=512

since the numerator and denominator were both divided by 2 hence the actual remainder is 512*2=1024.

Hi TG,

yesteday i was struggling with a problem which ccomes under remainder theory.Please explain as i tried to solve it with above given theoroms but resulted in a very complex solution.Just tell me if this problem could be solved in any simpler way.The problem is

3^1001 is devided by 1001 then what is the remainder.

Many thanks,

Nitin

@Shubhagata roy

rem of
7^115 divided by 114??

7^3=343

7^115=7^(3*38+1)

=>7^115=(343^38)*7

So, the qstn is now (343^38)*7/114.

Also,114*3=342.i.e 1 less than 343

Therefore by remainder theorem it becomes = (1^38)*7/114 =7/114

 =>Ans is 7

No fundoo bond

1001 is not a prime number.It is devisible by 7,11 and 13.

The answer is 947.

1001 = 77 * 13.

If we use chinese remainder theorem also,

a = 77, b = 13

r1 = 47, r2 = 9

x = -1, y = 6

Therefore remainder = axr2 + byr1 = -77*9 + 13*6*47 = 2973

Remainer (2973/1001) = 971.

So, i think 971 is the answer 

@ Deep Agrawal - for 7^115 / 114.

Phi(114)=36 and not 66.... .i.e. 114(1-1/2)*(1-1/3)*(1-1/19) = 36

therefore we are left with 7^7/114 and 7^3/114=1

=>7^7=7^6.7 =>qstn becomes 7/114= ans=7

Reminder(2^1990)divided by 1990

In this case N=1990 =2.5.199
so find the reminder when divided by 199 and multiply it wid 2*5 since 2 and 199 are co primes. again divide by 199 if its divisible.. thats the reminder..

suppose if we get 25 as reminder when divided by 199.

multiply 25 wid 2*5 ie 10.

we will get 250.. which should be divided by 199 agion to get the reminder..

this should be the simplified version of answer.

thanx fundooo for this mindblowing tutorial...it is of immense help...but i have a big doubt...does modular arithmetic follow these rules?:

1.(remainder of 1 expression)*(remainder of second expression)=(remainder of overall expression)

2..(remainder of 1 expression)+(remainder of second expression)=(remainder of overall expression)

3..(remainder of 1 expression)-(remainder of second expression)=(remainder of overall expression)

does this mean if 1998*1999*2000 is divided by 47,we find remainders individually with respective terms in the expression n multiply them to get the final answer??

@ rishu batra

Though I was not asked but still i felt like answering it...

Answer to ur qstn is a simple yes..

the rule says that all the operators (+,-,*) remain as they are and need to be operated on the individual remainders  which are obtained by separately finding out remainder for each of the numerators..

Hi

there is a error in d solution of ex. to find last 3 digit of 57^802 while calculating ©(1000)=1000(1-1/2)(1-1/4),it shud be 1000(1-1/2)(1-1/3) as 1000=(2^3)(5^3)..

bye n tk cr

hi all,

I find one problem while solving one of the papers.

Q. what will be the remainder if (2222^5555+5555^2222) is divided by 7 ?

plase give answer with complete solution...

@ Amit Chandaliya

Hi Amit, perhaps this has been discusse in the forum before also...

2222 = 7*317+3 and 5555 = 7*793+4

=> Qstn becomes (3^5555+4^2222)%7

3^3 % 7 = 27%7 = -1

4^3 % 7 = 64)%7 = 1

=>3^5555 = 3^(3*1851+2)

and 4^2222= 4^(3*740+2)

now, we have first part of qstn as :-

 ((27)^1851 * (3^2)) %7 = (-1)^1851 * 2 =-1*2 = -2

and second part is :-

((64)^740 * (4^2)) %7 = (-1)^740 * 2 =1*2 = 2

Therefore answer wud be (-2)+(2) = 0

Hope this is a full solution.....

 @love kumar

U found a mistake but u didn't look into it carefully.Its a typo(printing mistake)  not an error.

The typo is dat  Â©(1000)=1000(1-1/2)(1-1/4) shud be 1000(1-1/2)(1-1/5) as as 1000=(2^3)(5^3).. => u have to take 2 and 5 as "a" and "b" respectively.

The resultant value comes to 400 only...

so solution is alright...hope you got ur mistake....

hi all,

how do we solve the followin problem:

find the remainder when 123412341234.....(89times)is divided by 19?

hi varun!!

pretty rite..

thanks

As the theories given by fundoo and tG r fabulous. But I have some queries.

IN wilson’s theory,how the 2^nd example gives 1 remainder.pleae somebody make it clear. Treat this question as u have not done 1^st example of same theorem.

Also explain the Wilson’s theorem morefor my convenience.

@Ameya Raich

you are getting wrong in the very preliminary step.

which is 2^11 % 25= -1  => wrong buddy

infact it shud be 2^10 % 25= -1

Follow the very same steps that u have taken.....

You will get the very same answer in both the case and for ur ease i wud say the answer is 88

Also u said that 

Rem( (2^990)/ 25)
= Rem (2^11/25 * 90) =1.
Thus we are left with Rem (2^7/25) = 7; =>wrong,it shud be 3
now frm step (1) above ,
final answer is 4*7 = 28 .

I am rewriting the above steps with corrections

Rem( (2^990)/ 25)
= Rem (2^(10*99+7)/25) => (-1)^99 * rem (2^7 /25) => -3 =>22

final answer is 4*22 = 88

hi everybody.....

plz solve (19^36 + 17^36) %111...

hi shashank,

if this is the qn...

(777)^777/1001

1001=(7*11*13)

so basically u have to find remainder of

(777)^776/(11*13)

do this by chinese thm and then multiply the resultant rem by 7.

if u still need my assistance done be afraid to revert.

regards,

fundoo

hi varun ,

it can be solved using chinese remainder theorem

 .. (19^36 + 17^36) %111 so i will divide this into two part

fist 19^36%111 and other 17^36%111

so our  111 ==> 3*37 both are prime

so a=3  now we have to find the remainder for

19^36/3 =>1=r1

19^36/37=> 1=r2 so  and

our a and b are 3 and 37 so   3*x+37*y=1 it gives me x=-12 and y=1

so   3*-12*1+37*1*1=  -36+37 =>1

now same way it can be done for 17^36

so 17^36/3 => 2^36/3 => 8^12/3 => (-1)^12/3 =>1

17^36/37 =>1

so it will also follow the same pattern and answer will be 2 =>1+1...

find the last three digits of 3^{1994 .}

Now as per Euler theorem , Rem [3⁴⁰⁰/1000] = 1 .

Similarly , Rem  [3²⁰⁰⁰/1000] = 1 . Can anybody tell me how do we proceed further ? How do we handle the extra 3^{6 ?}  

Too good...Some of the problems are solved in Arun Sharma's book and it is always good to see alternative solutions to such kind of problems.

Regards,

Vipul

hey kamal  ,

Thanks for posting the solution ... but i am not aware of the Eulid's theorem ... can you please post that too ??

However , i have found one other way to solve the problem

3^{1994 =} 9⁹⁹⁷=( 10 - 1) ⁹⁹⁷ .

now using binomial expansion , we can get the laste three digits . the answer comes outtabe 369 .  

Just Amazing JOB.....fundoo.. fantastic...stuff.

Thanks,

HG

Hi vamshi

is the answer 6???

please confirm

Hi....One thing that I am not getting in Chinese theorm is how to set values for x,y because there might be the case when there are more than 2 values satisfying eq ax+by=1...for eg  what is the remainder when we divide 123456789 by 6.....

i am not getting the ans in any case...

how do you solve 38! / 41 ?

if we use the same method used to solve 39!/41, then it looks something like this.

40! / 41 = 40.39.38! / 41

Remainder [ 40! / 41 ] = Remainder [ 40.39.38! / 41]

ie. 40 = 40 . Remainder of (39*38! / 41)

that means remainder of (39.38! / 41) should be 1.

that means remainder of (38! / 41) will be 1/39 ??????

i am not convincedd.

please clarify??????

hi TG!!!

plz help me wid  dis problem - rem(8^182)/65

i ws doing it wid chinese remainder theorem-

65=13*5, >>>>>a=5,b=13;

rem(8^182)/5=rem((8^45))^4)*(8^2)/5)=4>>>>r1

rem(8^182)/13=12>>>r2

now 5x+13y=1;>>>x=-5,y=2;

so ar2x+br1y=212???????

i cud get da answer as follows-

phi(65)=48

rem((8^182)/65)=rem((8^48)^3)*8^38)/65

>>=rem(8^38)/65

>>rem((8^4)^9*8^2)/65

rem(8^4)/65=1

thrfre>>=rem(8^2)/65=64

but wat abt chinese remainder theorem wats wrong wid dat??

hi guys......this is an interesting question for all to work out...very simple one.....just a small trick involved....

if A=33*32*31*30*29*28

&

if Remainder [ A/(12k)] = 0......then find the minimum value of k?

really awesome fabulous and excellent ...

Thanxx Fundoo Bond..

u did a gr8 job..

Mind blowing

Hi,

The value of k = 1 for sure.

Bcoz 12 = 3 x 4 & we already have one 3 and one 4 in the numerator(ie, 33 = 3x11, 28=4x7).

So when A is divided by 12, surely the remainder will be zero.

Hence, K can be 1 since minimum value is asked.

Hi,

The value of k = 1 for sure.

Bcoz 12 = 3 x 4 & we already have one 3 and one 4 in the numerator(ie, 33 = 3x11, 28=4x7).

So when A is divided by 12, surely the remainder will be zero.

Hence, K can be 1 since minimum value is asked.

hi neo.u can find out the answer by the remainder theorm.the answer will be 64

good ,nice theorems ..its time saving a lot ..

please help me 2 solve this :

 find the 3 digit prime no. thats divides 2000!divided by 1000!^2

Hey,

The last example (while explaining the concepts i.e. Find the remainder when 8^643 is divided by 132 )that was solved had made use of Euler's theorum, whereas I tried to make use of Chinese remainder Theorum by the following method :

Break 132 into 4 and 33 and since 4 and 33 are co-primes therefore find 

Rem(8^643/4) = 0=r₁

Rem(8^643/33) -> Applying euler's theorum we get remainder as 17=r₂

Hence, Rem(8^643/132) = 4r₂x + 33r₁y where 4x+33y=1 (x,y) = (-8,1)

Hence, Rem(8^643/132) = 4*17*-8 + 33*0*1 = -544 , which is not the right solution.

Kindly help  and point out my mistake in the approach adopted.

Regards,

Pooja

hi sir...........im amit.....the way you are solving d problems is excellant...its rocking sir....                                                                 

Sir i didnt understand the Euler's theorem properly.....mainly the second line of the first example........plz explain again........

Hey FB

U truly are a bond with hell lot of fundas...

Thnx.

IPS.

Great work Boss..U r truly TG(total genius....)

Fantastic,Mindblowing!!!!!

hi...

could you please elaborate more in solving the above question..

(ii) remainder of 2 ^1990 / 1990
1990=2.5.199
from Euler's Theorem
phi(2)=1
phi(5)=4
phi(199)=198

////////// according to euler's theorm remainder of [2^phi(2) / 2] = 2^1 /2 comes out to be 1 but its zero..could u please correct me if m interpreting it wrong and please throw some light on the part right below..that LCM and mod thing ...unable to get it... ////

2^4=1 mod 5--------(1)
2^198=1 mod 199---------(2)
LCM(4,198)=396
so
2^396=1 mod 1990
2^1980=1 mod 1990(because 1980 is a multiple of396)
so now we remain with
2^10 mod 1990
1024 mod 1990=1024

Hi AD P,

Anwer to the (2^1990 % 1990) is impressive, but can you explain step LCM(4,198). Why did you take LCM ?

Sanjeev

thanks a lot.....!!!!!!!

It's really a great help ... for we don't find these sort of stuffs in books easily.... though a bit confusing at first .....

With so many theorems to solve similar Quants question .... it's a bit tough to judge which one to apply... without much practice ....

Thanks a lot!I always wondered how to answer such questions.This article was wonderful.do keep posting such articles.

I have a query.

In case of Chinnese method:

1. Do we need to check whether the number and divisor to be co-prime
2. If yes, can you explain me your last example 8^643 / 132 using Chinnese  method

Please explain

Cheers,
Sanjeev

Thank you for this beautiful article,

What is the remainder incase of 13^40 mod 49 ?

Cheers,

Sanjeev

hi, this is my first post to TG...

(2^2 + 22^2 + 222^2...(2222...48times)^2) % 9

[(4*(1^2) + 4(11^2) + 4(111^2)...] %9

4[1^2 + 11^ 2 + 111^ 2...] % 9

4[1 + (9+2)^2 + (108+3)^2...] % 9...(all terms of form (a+b)^2, where a is a multiple of 9, like 9, 108, 1107...)

4[1^2 + 2^2 + 3^2+...+48^2} % 9 .........(since all terms of form a^2+2ab are divisible by 9)

[4*48*49*97/6] % 9

remainder = 5...

please let me know if this is correct...

(13^40) % 49

this can be written as...

=[(2*7-1) ^40] % 49

By binomial expansion...

=(2*7)^40-40C1 * (2*7)^39 (1)...40C39 * (2*7)(-1)^39 + (-1)^40

all the operands in the above equation except for the last two are divisible by 49. So we need to find the remainder, when the last two terms are divided by 49.

=> -560 % 49 + 1

=7 * (40 * -2)%7 + 1

=7 * (-10) + 1

= -70 + 1

= -69

= 49*2 - 69

=29

Hence the remainder is 29....

Somebody please let me know if this is the right answer...

Thnxs TG.

Plz solve my problem.

Q.1 What is the remainder when (1!)^3 +(2!)^3 +(3!)^3 +(4!)^3 + -----------------------(1152!)^3 is divided by 1152??????

Q.2 What is the remainder when 2(8!) -21(6!) divides

          14(7!) + 14(13!)?????????????????????

Waiting for ur reply.

Ans 2:- 2(8!)-21(6!) = 2(8!)-3.7.(6!) or

                            = 2.8.(7!)-3.(7!) or

                            = 13(7!)

Now, remainder of 14(7!) + 14(13!) = remainder of 14(7!) + remainder of 14(13!) or

=14/13 + 0 or

=14/13 (ans.)

Q.1 What is the remainder when (1!)^3 +(2!)^3 +(3!)^3 +(4!)^3 + -----------------------(1152!)^3 is divided by 1152??????

Ans 1 :- 1152 = 2^7 . 3^2

Therefore, we can easily calculate the remainder of each term individually here.

e.g. all terms after (3!)^3 are completely divisible by 1152

hence, the remainder is 225 (which is equal to the remainder of the sum of first 3 terms).........Ans

Thnxs,

I ve another problem.

 Q .1 Find the 28383rd term of the series: 12345678910111112...........??????

hi tg,

QUES.find the last digit of the product of all 2 digit numbers that give a remainder of 2 when divided by 5?

hi tg sir,

i have a conceptual doubt.what if, in chinese reainder theorem,

N=a*b*c. That is N has 3 prime factors.can the chinese remainder theorem still be applied.if so then how?what about ax+by=1 in that case?

Hi Pushpender,

63=3^2*7

phi(63) = 63(1-1/7)(1-1/3) = 36*6/7*2/3 = 36

Thanks Sir for these tricks and examples..

Pls sir write as much as possible becoz it helps to us those who are unable to take or admit into your class room coaching.

Thanks Sir.

Really a very helpful and effective article .  An eye opener

I approach such problems, in which the numerator and the dinominator has some common factors, in the following way

1) take the common factors out i.e. divide both numerator and dinominator by their HCF

2)Find the remainder( now the numerator and the dinominator would be co-primes)

3) multiply the remainder with the HCF

So in this problem, we can write

2¹⁹⁹⁰ = 2* 2¹⁹⁸⁹

1990=2*5*199

phi(5) =4

phi(199) =198

LCM(4,198) =396

therefore 2³⁹⁶ = 1 mod 995

therefore remainder when 2^1989 divided by 995 is 2⁹

hence the required remainder is 2*2⁹

I think this clears a lot of cloud on the theory being applied here

I am new here...correct me if I am wrong

1001 is a beautiful number. any three digit number multiplied by 1001 results in the number written twicei.e. abc * 1001 =abcabc

in the prob 1111...2008 times is divided by 1001

111111 is divisible by 1001( 111 * 1001)

that means 111.....2004 times would be divisible by 1001

(111111* 1000001000001000....etc)

that would leave us with 1111

the remainder would be 110

@pravesh

I think the best approach in such cases is by taking LCMs, and then getting a smaller number to divide

Have a crack at this

remainder when 72! is divided by 73*36!

(I dont know where this problem appears. I got this from a friend through SMS. Hopefully I am not infringing any copyright )

Hi TG

i used some other method to solve 32^32^32 when divided by 9, by my method i am getting remainder =2 when 32^32^32 is divided by 9.

we can write the question as 2^5^32^32 where 5^32^32 can be wriiten as 3k +1 where k is an even number.

so now 2^3k+1 divided by 9 will give you 2 as the remainder.Could you please point out my mistake.

Hi Kulvir.

Thank you for the solution.I have a doubt which is:

(111..7) times whendivided by 7 will give a remainder 1 and if (111..14 )times will also give the remainder as 1 when divided by 7.?Is this correct.

Could you explain me the logic behind the chinese theorem. I mean how is it derived since thr has to be a logic behind...thanks 

Hi TG

Thanks...This site is 2 gud 2 b true!!1 more doubt!

#how many pairs of x and y exist for which the

eq: sqrt X + sqrt Y= Sqrt 1332 holds true.Could you plz solve it.It is one of the questions from TG quizzes.Are solutions given to the quizz problems on the site? Thanks

good article..cheers

can anyone help me with this??

find 28383rd term of the following 123456789101112...

options are 3,4,9,7..

my answer is coming 3..but its wrong

did it by taking 1-9-->9*1digits

                       10-99-->90*2=180digits

                        100-999-->900*3=2700digits

total=2889digits of 1,2 and 3 digit numbers.

so, 4digit numbers will have  28383-2889=25494digits

so the actual number will be 999+[25494/4]=7372

while dividing with 4 we get remainder 2 so the required digit is going to be 3 as we'll have 7 3 7 2 7 3 7 3...and so on

what am i doing wrong??

yeah its from Sharma..where the answer is 9..

here's another sum

p=1!+2*2!+3*3!+........+12*12!

what'll be the remainder when p+3 is divided by 13!?

my approach is this

for any number we can use the following

when we have 1!+2*2!+3*3!+.....+(n-1)(n-1)! this form and it is divided by n! the remainder is -1...

(eg of -ve remainder: if we have 23/6, we can express 23 as 6*4-1(6k-1)

                                 or we can express 23 as 6*3+5(6k1+5)  ) 

eg: as in 1!+2*2!+3*3! when divided by 4! will give 23/24...

 here remainder is 23 which we can write as 24k-1...so i write the remainder as -1...thus 23=4!-1

thus in the given sum p/13! will give remainder as -1 and 3/13! will give remainder 3..

thus the remainder should be 3-1=2....

i hope my method is ok..i don't have the answer to this.

if anyone gets a different answer to this or has a different method please do share...

well here x will be divisible by 440 if it is divisible by 4,11,10(440=4*11*10).

11's divisibility rule is (sum of odd digits)-(sum of even digits)=0/11k

here if we take 98 7's then we have the following

777777......98digits000+777

the number 77777....98digits000 is divisible by 11(49*7-49*7=0)

                                               is divisible by 4(ending with 3 0's)

                                               is divisible by 10.

so answer should be remainder of 777/440 =(777-440)=  337

is this the answer??

@ Gowtham Muthukkumaran Thirunavukkarasu 

with regard to ur solution dated 13 May,2009....

Could u pls explain the last step of ur answer i.e.

Remainder of  5*10^99/(9*98) - 5/(9*98)
= 0
How come?

@ Nabanshu Bhattacharjee

could you pls throw some more light on the LCM approach that you have mentioned in one of ur posts. There is only 1 example which everyone has repetitively explained for the LCM approach. Could u give me any other example wherein u hv used this approach.

@ all

or maybe someone else has other examples.

hi TG ,

its amazing work you are doing. in this world of money makers you are doing social service and that too when you know this is the only way your work is going to be applauded. Good job done TG.

keep up the good show and cheers!!!!!!!

thanks

Aashish Biala

Hi Mate. Kudos to you for putting up this article. It presents the requisite stuff in a comprehensible manner.

I wonder why you haven`t included the 'Method of Splitting the Divisor' in your text..

For instance consider this one (This is a question from the test series of one of the leading instiutes).....

Qn. N = 7777.....7 upto 2n+1 digits, n>3. Find the remainder when it`s divided by 1232.....?

Sol. 1232 = 7 x 11 x 16

N = 7p = 11q + 7 = 16r + 1 for some values of p,q,r.

Trying to solve 7p = 11q + 7... As 11q should be a multiple of 7, we try q = 0 and get p = 1.

So, Rem [N/77] = 7 i.e. 77p + 7 = 16q + 1 or 16q - 77p = 6

p = 2 and q = 10 satisfy it. Hence the least such number is 16(10) + 1 = 161.

So, N = 77(16)k + 161.

I failed to understand this solution i.e. why would it give the remainder.

Please make me understand it.....

Ans is option 1) 0 since

a^n + b^n + c^n +... is divisible by a+b+c+.... when n is odd.

hey ....please let me knw hw to do that???

[{6^11}^7] find the last two digit...

please let me knw hw to do dat wd euler's method ..or if nyoder method u have ..

please explain it......

Hey do it using chinese remainder theorem,

take n=1, then N = 7777777, which needs to be diveded by = 1232

1232 = 77*16

a = 77 and b = 16

N/77 => r1 = 7

N/16 => r2 = 1

R[N] = a*r2*x + b*r1*y >>>> 77x + 112y

ax+by = 1

77x + 16y = 1 >>>> x = 5 and y = -24

R[N] = -2303 which is equal to 1232*2 - 2303 = 161

=>7^7=7^6.7 =>qstn becomes 7/114= ans=7 can u xpalin dis step??