Number System 2: Remainders  
Suppose the numbers N _{1 }, N _{2 }, N _{3 }.. give quotients Q _{1 }, Q _{2 }, Q _{3 }.. and remainders R _{1 }, R _{2 }, R _{3 }..., respectively, when divided by a common divisor D. Therefore N _{1 }= D × Q _{1 }+ R _{1 }, N _{2 }= D × Q _{2 }+ R _{2 }, N _{3 }= D × Q _{3 }+ R _{3 }.. and so on. Let P be the product of N _{1 }, N _{2 }, N _{3 }... Therefore, P = N _{1 }N _{2 }N _{3 }.. = (D × Q _{1 }+ R _{1 })(D × Q _{2 }+ R _{2 })(D × Q _{3 }+ R _{3 }).. = D × K + R _{1 }R _{2 }R _{3 }... where K is some number  (1) In the above equation, only the product R _{1 }R _{2 }R _{3 }... is free of D, therefore the remainder when P is divided by D is the remainder when the product R _{1 }R _{2 }R _{3 }... is divided by D. Let S be the sum of N _{1 }, N _{2 }, N _{3 }... Therefore, S = (N _{1 }) + (N _{2 }) + (N _{3 }) +... = (D × Q _{1 }+ R _{1 }) + (D × Q _{2 }+ R _{2 }) + (D × Q _{3 }+ R _{3 }).. = D Ã— K + R _{1 }+ R _{2 }+ R _{3 }... where K is some number (2) Hence the remainder when S is divided by D is the remainder when R _{1 }+ R _{2 }+ R _{3 }is divided by D. Examples:
Answer: the remainders when 1998, 1999, and 2000 are divided by 7 are 3, 4, and 5 respectively. Hence the final remainder is the remainder when the product 3 × 4 × 5 = 60 is divided by 7. Answer = 4
2 ^{2004 }is again a product (2 × 2 × 2... (2004 times)). Since 2 is a number less than 7 we try to convert the product into product of numbers higher than 7. Notice that 8 = 2 × 2 × 2. Therefore we convert the product in the following manner 2 ^{2004 }= 8 ^{668 }= 8 × 8 × 8... (668 times). The remainder when 8 is divided by 7 is 1. Hence the remainder when 8 ^{668 }is divided by 7 is the remainder obtained when the product 1 × 1 × 1... is divided by 7 Answer = 1
This problem is like the previous one, except that 2006 is not an exact multiple of 3 so we cannot convert it completely into the form 8 ^{x }. We will write it in following manner 2 ^{2006 }= 8 ^{668 }× 4. Now, 8 ^{668 }gives the remainder 1 when divided by 7 as we have seen in the previous problem. And 4 gives a remainder of 4 only when divided by 7. Hence the remainder when 2 ^{2006 }is divided by 7 is the remainder when the product 1 × 4 is divided by 7. Answer = 4
Again 25 ^{25 }= (18 + 7) ^{25 }= (18 + 7)(18 + 7)...25 times = 18K + 7 ^{25 } Hence remainder when 25 ^{25 }is divided by 9 is the remainder when 7 ^{25 }is divided by 9. Now 7 ^{25 }= 7 ^{3 }× 7 ^{3 }× 7 ^{3 }.. (8 times) × 7 = 343 × 343 × 343... (8 times) × 7. The remainder when 343 is divided by 9 is 1 and the remainder when 7 is divided by 9 is 7. Hence the remainder when 7 ^{25 }is divided by 9 is the remainder we obtain when the product 1 × 1 × 1... (8 times) × 7 is divided by 9. The remainder is 7 in this case. Hence the remainder when 25 ^{25 }is divided by 9 is 7. Some Special Cases: 2.1A When both the dividend and the divisor have a factor in common. Let N be a number and Q and R be the quotient and the remainder when N is divided by the divisor D. Hence, N = Q × D + R. Let N = k × A and D = k × B where k is the HCF of N and D and k > 1. Hence kA = Q × kB + R. Let Q _{1 }and R _{1 }be the quotient and the remainder when A is divided by B. Hence A = B × Q _{1 }+ R _{1 }. Putting the value of A in the previous equation and comparing we get k(B × Q _{1 }+ R _{1 }) = Q × kB + R > R = kR _{1 }. Hence to find the remainder when both the dividend and the divisor have a factor in common,
Examples
The common factor between 2 ^{96 }and 96 is 32 = 2 ^{5 }. Removing 32 from the dividend and the divisor we get the numbers 2 ^{91 }and 3 respectively. The remainder when 2 ^{91 }is divided by 3 is 2. Hence the real remainder will be 2 multiplied by common factor 32. Answer = 64 2.1B THE CONCEPT OF NEGATIVE REMAINDER 15 = 16 × 0 + 15 or 15 = 16 × 1  1. The remainder when 15 is divided by 16 is 15 the first case and 1 in the second case. Hence, the remainder when 15 is divided by 16 is 15 or 1. > When a number N < D gives a remainder R (= N) when divided by D, it gives a negative remainder of R  D. For example, when a number gives a remainder of 2 with 23, it means that the number gives a remainder of 23  2 = 21 with 23. EXAMPLE
Answer: 7 ^{52 }= (7 ^{4 }) ^{13 }= (2401) ^{13 }= (2402  1) ^{13 }= 2402K + (1) ^{13 }= 2402K  1. Hence, the remainder when 7 ^{52 }is divided by 2402 is equal to 1 or 2402  1 = 2401. Answer: 2401. 2 .1C When dividend is of the form a ^{n }+ b ^{n }or a ^{n } b ^{n }:
EXAMPLES
Answer: The dividend is in the form a ^{x }+ b ^{y }. We need to change it into the form a ^{n }+ b ^{n }. 3 ^{444 }+ 4 ^{333 }= (3 ^{4 }) ^{111 }+ (4 ^{3 }) ^{111 }. Now (3 ^{4 }) ^{111 }+ (4 ^{3 }) ^{111 }will be divisible by 3 ^{4 }+ 4 ^{3 }= 81 + 64 = 145. Since the number is divisible by 145 it will certainly be divisible by 5. Hence, the remainder is 0.
Answer: The remainders when 5555 and 2222 are divided by 7 are 4 and 3 respectively. Hence, the problem reduces to finding the remainder when (4) ^{2222 }+ (3) ^{5555 }is divided by 7. Now (4) ^{2222 }+ (3) ^{5555 }= (4 ^{2 }) ^{1111 }+ (3 ^{5 }) ^{1111 }= (16) ^{1111 }+ (243) ^{1111 }. Now (16) ^{1111 }+ (243) ^{1111 }is divisible by 16 + 243 or it is divisible by 259, which is a multiple of 7. Hence the remainder when (5555) ^{2222 }+ (2222) ^{5555 }is divided by 7 is zero.
(a) 317 (b) 323 (c) 253 (d) 91 Solution: 20 ^{2004 }+ 16 ^{2004 } 3 ^{2004 } 1 = (20 ^{2004 } 3 ^{2004 }) + (16 ^{2004 } 1 ^{2004 }). Now 20 ^{2004 } 3 ^{2004 }is divisible by 17 (Theorem 3) and 16 ^{2004 } 1 ^{2004 }is divisible by 17 (Theorem 2). Hence the complete expression is divisible by 17. 20 ^{2004 }+ 16 ^{2004 } 3 ^{2004 } 1 = (20 ^{2004 } 1 ^{2004 }) + (16 ^{2004 } 3 ^{2004 }). Now 20 ^{2004 } 1 ^{2004 }is divisible by 19 (Theorem 3) and 16 ^{2004 } 3 ^{2004 }is divisible by 19 (Theorem 2). Hence the complete expression is also divisible by 19. Hence the complete expression is divisible by 17 × 19 = 323. 2.1D When f(x) = a + bx + cx ^{2 }+ dx ^{3 }+... is divided by x  a
EXAMPLES
Answer: The remainder when the expression is divided by (x  (1)) will be f(1). Remainder = (1) ^{3 }+ 2(1) ^{2 }+ 5(1) + 3 = 1
Since the expression is divisible by x  1, the remainder f(1) should be equal to zero. Or 2  3 + 4 + c = 0, or c = 3. 2 .1E Fermat's Theorem
EXAMPLE
Answer: Since 7 is prime, n ^{7 } n is divisible by 7. n ^{7 } n = n(n ^{6 } 1) = n (n + 1)(n  1)(n ^{4 }+ n ^{2 }+ 1) Now (n  1)(n)(n + 1) is divisible by 3! = 6 Hence n ^{7 } n is divisible by 6 x 7 = 42. Hence the remainder is 0. 2.1F Wilson's Theorem
EXAMPLE
16! = (16! + 1) 1 = (16! + 1) + 16  17 Every term except 16 is divisible by 17 in the above expression. Hence the remainder = the remainder obtained when 16 is divided by 17 = 16 Answer = 16 2.1G TO FIND THE NUMBER OF NUMBERS, THAT ARE LESS THAN OR EQUAL TO A CERTAIN NATURAL NUMBER N, AND THAT ARE DIVISIBLE BY A CERTAIN INTEGER To find the number of numbers, less than or equal to n, and that are divisible by a certain integer p, we divide n by p. The quotient of the division gives us the number of numbers divisible by p and less than or equal to n. EXAMPLE 14. How many numbers less than 400 are divisible by 12? Answer: Dividing 400 by 12, we get the quotient as 33. Hence the number of numbers that are below 400 and divisible by 12 is 33. 15. How many numbers between 1 and 400, both included, are not divisible either by 3 or 5? Answer: We first find the numbers that are divisible by 3 or 5. Dividing 400 by 3 and 5, we get the quotients as 133 and 80 respectively. Among these numbers divisible by 3 and 5, there are also numbers which are divisible both by 3 and 5 i.e. divisible by 3 x 5 = 15. We have counted these numbers twice. Dividing 400 by 15, we get the quotient as 26. Hence the number divisible by 3 or 5 = 133 + 80  26 = 187 Hence, the numbers not divisible by 3 or 5 are = 400  187 = 213. 16. How many numbers between 1 and 1200, both included, are not divisible by any of the numbers 2, 3 and 5? Answer: as in the previous example, we first find the number of numbers divisible by 2, 3, or 5. from set theory we have n(AUBUC) = n(A) + n(B) + n(C)  n(A intersectn B)  n(B intersectn C)  n(A intersectn C) + n(A intersectn B intersectn C) n(2U3U5) = n(2) + n(3) + n(5)  n(6)  n(15)  n(10) + n(30) > n(2U3U5) = 600 + 400 + 240  200  80  120 + 40 = 880 Hence number of numbers not divisible by any of the numbers 2, 3, and 5 = 1200  880 = 320. I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on this in the CBT Club this week.

Re: Number System 2: Remainders  
hi, this is a good one article. piyush 
Re: Number System 2: Remainders  
Hi TG, Thank you for this is excellent article for starters. Regards, Nikhil

Re: Number System 2: Remainders  
if R=[(0.15)^{55}+(0.25)^{55}]/[(0.15)^{54}+(0.25)^{54}] 1.R>0.20 2.R>0.1 3.R=4 4.R<0.25 I am new here plz reply...... 
Number System 2: Remainders  
Can anyone help in finding the solution of a problem. The problem is find the remainder when 444^444^444/7. 
Re: Number System 2: Remainders  
hi can anybody brief it .i cudnt get it

Re: Number System 2: Remainders  
Nice one......better than any cat prep book i have consulted..... doing grt job sir... 
Re: Number System 2: Remainders  
[3^101]/77 find the reminder? 
Re: Number System 2: Remainders  
[17^36+19^36]/111 what is the reminder? 
Re: Number System 2: Remainders  
Got the ans of 3^101 /77 . ans is 47. 
Re: Number System 2: Remainders  
@venkat 49 and 288. n.(n+1)/2 = x^2 => n/2 and (n +1) both are perfect squares. So look for an odd square and half of its next number should also be a square 
Re: Number System 2: Remainders  
hi sir i wann aknow in ques 16 why u hav added 80 instead of subtracting it?????

Re: Number System 2: Remainders  
@ hardik.. is the ans for your ques is 52.. do let me knoe 
Re: Number System 2: Remainders  
hi varun, how to check : AB36 (A & B is any no. ) is perfect square this type of questions please provide solution 
Re: Number System 2: Remainders  
hardik..how u got answer.....
Got the ans of 3^101 /77 . ans is 47. could you please explain 
Re: Number System 2: Remainders  
Hi Cn you plz tel hw did u gt 47. ans shud be 26. please explain. 
Re: Number System 2: Remainders  
sorry guys i'v got 47.i was doing it in a diff. way 
Re: Number System 2: Remainders  
better method (444^444)/7=(3^444)/7=(9)^222/7=(2)^222/7=(8)^74/7=1 best of luck (soln courtesy to my roomate at IIT kgp hostel abhisek) 
Re: Number System 2: Remainders  
what is the remainder of 2^{147 }divided by 11 ??? can anybody give solution how to solve this. 
Re: Number System 2: Remainders  
2^{10} = 1 mod11 => 2^{147} = 2^{14x10 + 7} = 2^{7} mod11 = 128 mod11 = ^{}7 mod11^{ } 
Re: Number System 2: Remainders  
Given that N = aD + 4 9N = bD + 7 => D > 7 from eqn 1, we get 9N = 9aD + 36 comparing with eqn 2 36 = cD + 7 => cD = 36  7 = 29 => D = 29 as it cannot be 1. 
Re: Number System 2: Remainders  
it is not clear kamal.... how it came...? 36 = cD + 7 
Number System 2: Remainders  
hi evryone...pls help me solve this problem. find the remainder when 1!+2!+3!+........+49! is divided by 7. thanks 
Re: Number System 2: Remainders  
find the remainder for ( 5^99) / 66 
Re: Number System 2: Remainders  
thanks saloni..but what if the divisor is 17 instead of 7.??is there any shortcut?? 
Re: Number System 2: Remainders  
Ø(66) = 66(1  1/2)(1  1/33) = 32. => 5^{32} = 1 mod66. => 5^{99} = 5^{32x3 + 3} = (5^{32})^{3} x 5^{3} = 5^{3} mod66 = 125 mod66 = 59 mod66. 
Re: Number System 2: Remainders  
@ ravi : I don't know of any shortcut. If you find any shortcut , Please do let me know. 
Re: Number System 2: Remainders  
can ny1 tell me where r the new posts by TG..m new to this site.. 
Re: Number System 2: Remainders  
are u sure ur answer is correct kamal..Can u elaborate ur technique.didnt get it. 
Re: Number System 2: Remainders  
Reckon Ø(66) has to be 66(1  1/2)(1  1/3)(1  1/11) ? 
Re: Number System 2: Remainders  
[17^36+19^36]/111 what is the reminder? 
Re: Number System 2: Remainders  
What is the remainder of the 2000^1000 when divided by 13 
Re: Number System 2: Remainders  
3 is the answer 
Re: Number System 2: Remainders  
8927 digits starts with 6 any two digit number is divisible by 17 or 23.what will be the 5293rd digit ? 
Re: Number System 2: Remainders  
thank you sourav it was a gr8 help 
Re: Number System 2: Remainders  
Namit, 16 is a factor of 10^4, therefore divide the last four digits of N to get the remainder. Answer = 1000 mod 16 = 8. Refer Divisibility and Bases for more details.  SE 
Re: Number System 2: Remainders  
hi namit.... the answer is 13 the last four digits of d no formed by the first 1000 digits is 3693 which divided by 16 gives a remainder of 13..... 
Re: Number System 2: Remainders  
sourav can you tell me how 1099 you are getting 100numbers in my calculations it is 93 numbers 
Re: Number System 2: Remainders  
hello TG sir can you please tell me answers of the above questions they are toubling me a lot ..... 
Re: Number System 2: Remainders  
q> 232 to 756 how many 5 ,s are there ? 
Re: Number System 2: Remainders  
Oh! "number formed by the first thousand digits (from left) of N" and I divided the last four digits of N. 
Re: Number System 2: Remainders  
please solve this ques.. What is the remainder when 6^{83 }+ 8^{83 }is divided by 49 ?? 
Re: Number System 2: Remainders  
now solve 1 7^7^7 mod 13 2 25^25^25 mod 9 3 37 ^47^ 57 mod 16 
Re: Number System 2: Remainders  
Hi all ! A good one ! Proved quite helpful to me ! 
Re: Number System 2: Remainders  
plz solve Remainder when 1^39 + 2^39 + 3^ 39 +...+ 12^39 is divided by 39. 
Re: Number System 2: Remainders  
plz tell me how to solve follwing, 21^875/17 what is the reminder? piyush 
Re: Number System 2: Remainders  
hardik,(late again) the answer of [3^101]/77 is 47 
Re: Number System 2: Remainders  
Hi All
pls give me sol of following ques.Ans is 7 How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers 
Re: Number System 2: Remainders  
Hi Shivam /all pls tell me following ques , its answer is 320. Let f be a factor of 120, then the number of positive integral solutions of xyz = f is 
Re: Number System 2: Remainders  
can anyone plz tell me the answer wid Solution..? 6^83+8^83/49 (6 Raise to powe 83) + (8 Raise to powr 83) Divided By 49 rEMAINDER? 
Re: Number System 2: Remainders  
Hi pallav, 6^83+8^83/49 phi(49)=42 ==> (6^42+8^42)%49=1 Now, (6^83+8^83)%49=(6^1 +8^1)%49= 14 14+49===>35(Remainder) 
Re: Number System 2: Remainders  
HI sir, nice article 
Re: Number System 2: Remainders  
very awesome reply brthr 
Re: Number System 2: Remainders  
Hi abhi, (1!)^{3}+(2!)^{3}+(3!)^{3}+.............+(1152!)^{3}/(1152!) is the remainder 0? will post approach if answer is confirmed.. Rohit 
Re: Number System 2: Remainders  
hiii rohit gud try but dude i want 2 say dat me sorry its 1152 in reamiander instead of 1152! n answer is 225 . cheers 
Re: Number System 2: Remainders  
probably better soln: 2^147/11 = (2^5)^29*4/11 = (1)29*4/11 = 4/11 Final Remainder = 114 = 7 
Re: Number System 2: Remainders  
hi atishay can u pls tell me hw u found the last 4 digits of that number?

Re: Number System 2: Remainders  
Hi, Please tell method to solve what is the remainder when 90^{91 }_{ is divided by 15 (as 15 and 90 are not co prime)Thanks in Advance} 
Re: Number System 2: Remainders  
I think its remainder is 0 
Re: Number System 2: Remainders  
Hi Pankaj ans is 0. 
Re: Number System 2: Remainders  
Dear TG Sir, Thank you very much.

Re: Number System 2: Remainders  
Hi, Can any1 explain this ???? 
Re: Number System 2: Remainders  
Hello Sir can u explain me Fermat theorem more broadly. 
Re: Number System 2: Remainders  
Brilliantly complied and beautifully explained !! Thanks TG 
Re: Number System 2: Remainders  
please can u explain this solution 
Re: Number System 2: Remainders  
can someone help me in solving this find the 28383rd term of series: 1234567891011121314.... Ans: 3 
Re: Number System 2: Remainders  
plz solve this prob: what is the remainder when(999/99)^9? plz state the method and the logic too..finding too much diff in solving this... 
Re: Number System 2: Remainders  
Hi Shubham Please rewrite the question clearly that which number is being divided by whom as there can be many (mis)interpretations of your statement. Kamal Lohia 
Re: Number System 2: Remainders  
find the remainder. 3^5^7^9 divided by 41. 
Re: Number System 2: Remainders  
also find the remainder of 5^25^125^3125 divided by 11. 
Re: Number System 2: Remainders  
Thanks for yr help....as a matter of fact I tried this problem by using eruler equation but couldn't reached final answer......anyways keep doing great work...!!! 
Re: Number System 2: Remainders  
[17^36+19^36]/111 what is the reminder? Is the answer to the above problem is 2 ? 
Re: Number System 2: Remainders  
Debashish, Yes the remainder will be 2. 111 = 37 * 3 Φ(111) = LCM[Φ(37) ,Φ(3)]= 36 17^{36} mod 111 = 1 19^{36} mod 111 = 1 Hence 1 + 1 = 2 
Re: Number System 2: Remainders  
Absolutely Brilliant Article ! Thanks a lot! It covers so many questions asked in CAT or all entrance exams for MBA! Great Work! 
Re: Number System 2: Remainders  
Doubt : Frame a digit using only 1 and 0 divisible by 2011. 
Re: Number System 2: Remainders  
11111....._{2010 digits} is divisible by 2011. Now you can put any zeroes after this and number will remain divisible by 2011. Kamal Lohia 
Re: Number System 2: Remainders  
Awesome thanks for the explaination sir. 
Re: Number System 2: Remainders  
Hi Kamal thnks alot for reply!!! but please explain me in simple words... i still didnt get how to find remainders in factorials... 
Re: Number System 2: Remainders  
hey...can nybdy tell me were to get d solutions for d quize??????? plz help me out!!!!!!!!! thanx in advance!!!!!!! 
Re: Number System 2: Remainders  
nice....... 
Re: Number System 2: Remainders  
literally awesome.... 
Re: Number System 2: Remainders  
Your answer is correct ,but i guess the no would be 7373. (as the first no of the series will be 1000. so 1000 + 6373th no = 7373). Now, its 2nd digit will be 3. 
Re: Number System 2: Remainders  
Thank you sir but my question is that how do we proceed in questions where two such numbers are not co prime. In this case 9 and 225. Thanks for the solution though 
Re: Number System 2: Remainders  
hi kamal, phi(66) = 20 not 32. so i think we'll have to do it again. 
Re: Number System 2: Remainders  
Hi Pranjal You are correct. Thanks for correction. Now solve the problem. Kamal Lohia 
Re: Number System 2: Remainders  
Hi 10 and 225 are not coprime so your first step altogether is incorrect when you write that 10^{90}/225 = 1. Check it. 
Re: Number System 2: Remainders  
2^147 / 11 =2^2 * 2^145 / 11 =4 * (2^5)^29 / 11 =4 * (32)^29 / 11 =4 * (1)^29 / 11 =4 Remainder = 114=7 
Re: Number System 2: Remainders  
Hi Sid, 20^1000 * 100^1000 / 13 =7^1000 * 9^1000 =(7^2)^500 * (9^2)^500 =(3)^500 * (3)^500 =3^1000 =3 * 3^999 =3 * (3^3)^333 =3 * (1)^333 =3 
Re: Number System 2: Remainders  
Thanks Sir for the reply. However, I am not been able to get the below statement  i.e. 104 = b*D + 23 or D Cn u pls explain this again? 
Re: Number System 2: Remainders  
Awesum !! Thanks ! 
Re: Number System 2: Remainders  
this is very good article of reminders in numbers thanx for it 
Re: Number System 2: Remainders  
Sir, Please help me in finding the answer. Question is : whats the remainder if you will divide 7^0+7^1+7^2+7^3+7^4......+7^365 with 5. Thanks in advance. 
Re: Number System 2: Remainders  
Is it 3? 
Re: Number System 2: Remainders  
but how. can u explain? 
Re: Number System 2: Remainders  
This is super awesome thank you 
Re: Number System 2: Remainders  
hi.. help me in dis questions.. what is the remainder of : (3^5^7^9)/41

Re: Number System 2: Remainders  
What is the remainder when 6^83 + 8^83 is divided by 49 ?? pie(49)=42 6^(841)+8^(841) so when (6^83+8^83)%49 = 68 = 14 =35(i.e;4914) "a^pie(N)%N = 1" 
Re: Number System 2: Remainders  
yes 47 is the answer to the que 3^101/77 
Re: Number System 2: Remainders  
a very nice article sir can u tell me the ans of 32^32^32 remainder when divided by 3,5,7,9,6 
Re: Number System 2: Remainders  
Hi Kamal Sir, I too have similar doubt. How 36= cD+7 Pl clarify..
Thanks, Reeti 
Re: Number System 2: Remainders  
Hi, I think i got the solution to it: 9aD+36=9N also, bD+7=9N Therefore, 9aDbD+367=0 Let, 9aDbD=cD so, 36=cD+7 cd=29 Pl, correct me if i m wrong!
Thanks, Reeti.

Re: Number System 2: Remainders  
No Reeti/Radha There is no need to correct you. You are absolutely right. Kamal Lohia 
Re: Number System 2: Remainders  
hi lonely planet, Q is 19 here..28/19= rem 9, 112/19=rem 17 
Re: Number System 2: Remainders  
Can you please explain, how u got to the answer? 
Re: Number System 2: Remainders  
use hit and trial....start from 10 as Q...u will stop at 19!!!!...take N corresponding to the Q!!..hope u got it!! 
Re: Number System 2: Remainders  
alrght man!!..but i think in dis case if we will go by the normal approach, time wud be consumed much!! 
Re: Number System 2: Remainders  
thank you sir!!!!... 
Re: Number System 2: Remainders  
gr8!!!! Thanks. 
Re: Number System 2: Remainders  
Thank u sir... gr8 approach... 
Re: Number System 2: Remainders  
Thanks a lot sir!!!
Regards, Radha

Re: Number System 2: Remainders  
Hi radha, Ans has to be none of these. As max value of elements is 121. I hope the ans is correct. Let me knw. Thanks n Regards, 
Re: Number System 2: Remainders  
The expansion got a fault 1^39+2^39+3^39+....+12^39 != (1+2+3+...+12)(1^38+2^38+....12^38)
Try for some small no.you will get it. 
Re: Number System 2: Remainders  
Dear TG, Help me solve the following., Find the reminder when (10^10)+(10^100)+(10^1000)+.....+(10^10000000000) is divided by 7. 
Re: Number System 2: Remainders  
Hi Rohit, http://totalgadha.com/mod/forum/discuss.php?d=1369 It is explained in the above lesson. Regards, GM 
Re: Number System 2: Remainders  
can some one help me solving this prob 128^1000/153 
Re: Number System 2: Remainders  
thanks for rply amresh....bt could you explain this step (106)^4 became 52 in nxt step did you do it by normal method of multiplying or any short cut???? 
Re: Number System 2: Remainders  
(10^10)/7=3*(((3)^3)^3)/7=(27^3)*3/7=((281)^3)*3=3/7 similarly other 9 and we get 30/7=5 
Re: Number System 2: Remainders  
ques Determine number of terms in expansion of (1+a1+a2+a3)^8???? 
Re: Number System 2: Remainders  
hi abhinav......there is mistake..... 4^4=64.........????????????4^3 is 64 i.e (4^3)148 now the answer will be 1. 
Re: Number System 2: Remainders  
1st of all segregate the terms of 444^444^444/7 as 444^444/7. from here we will get the remainder as 1, now 444^1/7=3 ans. 
Re: Number System 2: Remainders  
56 is the answer 
Re: Number System 2: Remainders  
What is the remainder of 59^73^5! when divided by 37? 
Re: Number System 2: Remainders  
can anyone please repost number system 1 forum.. i really need it plzzz 
Re: Number System 2: Remainders  
E(37)=36 59^73===59^36k+1 = 59^36k x 59 = 1x22=22 
Re: Number System 2: Remainders  
(6^17+17^6)/7=? 
Re: Number System 2: Remainders  
Better find the last two digit of the individual number and multiply it...Is the answer 25? 
Re: Number System 2: Remainders  
N = 33 D = 29 R[N/D] = 4 R[9N/D] = 9*R[N/D] = 9*4 = 36 Now, R[36/D] = 7 So, D = 26 & N = 33 
Re: Number System 2: Remainders  
53 
Re: Number System 2: Remainders  
Hi. Could anyone help me in solving this question. 3^101/77 
Re: Number System 2: Remainders  
hey,how u took common ,huh. 
Re: Number System 2: Remainders  
How 17^36 = 1 (mod37) please explain 