Remainders  
It is uncanny how children pick up a lot of small
habits and beliefs of their parents. Even the ones that rebel against their
parents bear subconscious resemblance to their father or mother. There is a
lesson for instructors in this. It is important for them to realize that students
mirror their feelings about CAT. If the instructor expresses or feels that CAT
is tough, or is fearful about CAT, his students will mirror the same feeling
and will be less confident. If the instructor is brazen and casual about the
paper and scoffs the competition, his students would reflect the same feelings.
Also, it is so necessary to have unflinching faith in one’s students. I still
remember that during my school days my mother used to proudly proclaim that I
was an intelligent kid. I was barely scrapping passing marks in the school
exams. If truth be told I was at the bottom of the class, but my mother had her
blinkers on. And because of my mother I also believed that I was second to
none. It was only years later, during my boards exams, that I took to studying seriously,
and managed to outperform everyone else. I don’t know if it was my mother’s
blind love for me or that she could see some bright spark in me that made her
claim my intelligence but it really had great effect on my attitude. And
attitude, in an exam like CAT, is everything.
While I am trying to write a DI lesson for the CAT CBT Club, here is another sparkling gem produced by a TGite already famous amongst you all Software Engineer. When I saw the sheer size of the article I was awed. And so will you all be. The hard work demands applause from all of you. So read the article guys and do not forget to thank Software Engineer for this one.  Total Gadha Join our CBT Club for further discussions. We shall cover some problems based on this in the CBT Club this week.

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Mind blowing article...Thx a ton SE Bhai... Keep rockin Ashwin 
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The patience has been rewarded!!! Once again a master piece! 
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Whoa ........ Such a compact article ...... Well if u divide this article, remainder will be 0 ...... it covers all, nothing left...... Gud work.... 
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vow......... Thanks a lot. 
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This is really a fantastic effort to help bell the CAT. Thanks S.E.Bhai 
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That it has taken TG sir 3 days to read and approve this article speaks volumes about this "gem of an article". Will update my grey cells thanking SE and TG all the time. 
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A piece of 'JEWEL' Thanks SE.. Reg, Ishaan 
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Indeed a master piece, as always. 
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SE Thanx a ton for ur efforts Havent gone in detail but definetely know it has to be amongst the best as it has always been Dude plz keep the good work goin 
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SE Bhai, you are gem of a person. Marvellous article indeed. 
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Sipra, Yes, it's correct. 
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Thank you! its really a very helpful article.. feel like I can solve any question on remainders now 
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what an article sir ji... thanx a ton...!! regards, Sridip. 
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just beautiful 
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nice Article. thanks SE Regards Amit 
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this really helps..... thanx a lot sirjeeeeeee. keepupdating on these type of articles..... 
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Great work. Thanks SE . 
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Ronak, Sipra, That's another typing mistake. : ( Correct one: 20 mod 100 Hence, answer is 20.  SE 
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FaNtAbUlOuS... 
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777^777 are the last 3 digits 973?? 
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Great artice SE sir.. this is going to be very helpful for everyone. Regards, The Killer 
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very nice article for beginners like me......... HOW 55^190=35mod153 came to 55^190=120mod153????? can't it be 55^190=118mod153? please help me in this.. 
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hello i cannot get the first three calculations.as to what is modulo. n how modulo order of 13 is 2.plz help me out 
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mind blowing article...thnx very much bhai expecting few more b4 the DDay... nice work... 
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When you write, " the remainder when b^{k }is divided by n, is same as the unit digit of b^{k} in base k" you actually mean b^{k} in base n, dont you SE? Is this a typo ? 
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Thank you very very much SE..You have made our path to IIMs much easier.. 
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Hey SE, Good work Also look at one Q'n...i.e. 22^1352 mod 52 is 16 not 14...check it out! HG 
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Dear SE, Thanks a lot for this article... it has been a great help for all of us.. now i am sure i will be able to crack remainder questions easily... 
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Kunal, Find the last three digits of b^K means Find the remainder when b^K is divided by 1000.  SE 
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SE, Thnx for ur post That i got it.........but how lemda(1000)= 100 bcas i m getting less than that......I am missing sth and i am nt able correct it.......pls help me out. Kunal 
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Kunal, 1000=8*125. HCF[8, 125]=1. L(1000) = LCM[L(8), L(125)] = LCM[2, 100] = 100.  SE 
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SE, I got it..............Thank you veru much again for your prompt reply. 
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Subhadip, In the problem of 2^2004 mod 2004,the final equation comes of the form R=4x=501y + 88, if we make it 4x  88 = 501y. Now LHS must be a multiple of 501, so x=275, 4x  88 = 4*275  88 = 1100  88 = 1012 (this is not of the form 501y) I think x=523. R= 4*523 mod 2004 = 2092 mod 2004 = 88 mod 2004  SE 
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great stuff man!! thanks 
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solve dis: Find remainder when (100^3+101^3+102^3............................+198^3) is divided by 9. 
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Hi Sergey, Q. Find remainder when (100^3+101^3+102^3............................+198^3) is divided by 9. A. Remainder is 0. 100+101+..+198=14751 which is completely divisible by the dividend as the powers of the dividend are odd. Also, 14751 is divisible by 9 (digital sum), Hence the remainder when (100^3+101^3+102^3............................+198^3) is divided by 9 is 0. Please check if the answer is correct. Hope this helps.. Rohit 
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Rohit, can u plz explain this step:"100+101+..+198=14751 which is completely divisible by the dividend as the powers of the dividend are odd.Since the powers are odd,can we aply dis fr all odd powers.Consider dis:2^5+3^5+4^5=1299 dis should be divisble by 9 as (2+3+4=9)but it is not.Kindly explain ur solution as i may nt be understanding ur solution.Although the solution u gave,gives the right answer. I gt one solution which says:sum of cubes of 3 cosecutive terms is always divisible by 9.hence the above terms make 33 such pairs and so gives the remainder zero. Can anybody explain sum of cube of 3 consecutive terms funda. 
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SE cn u help 
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Sum of three consecutive no.s= (a1)^3 + a^3 + (a+1)^3 = a^3  3a^2 + 3a 1 + 3a^3 + a^3 + 3a^2 + 3a + 1 = 3a^3 + 6a = 3a(a^2+2) Now if a is a multiple of 3 the this whole becomes a multiple pf 3*3 that's 9 If a= 3n+1 then 3(3n+1)(9n^2 + 6n + 1 + 2) = 3(3n+1)(9n^2+6n+3) =9(3n+1)(3n^2+2n+1) which is again a multiple of 9 If a= 3n+2 then 3(3n+2)(9n^2 + 12n + 4 + 2) = 3(3n+2)(9n^2+12n+6)= 9(3n+2)(3n^2+4n+2) again a multiple of 9 so 3a(a^2+2) is always a multiple of 9 or sum of three consecutive no.s is always a multiple of 9 
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Hi sergey.. Sorry for the late reply..was busy lately and also did not check this thread..that's a very good point u raised(may b i have goofed up in the concept )..but if it is checked with other numbers it becomes true..i will try to find out the exact reason/concept and let u know about it. (TG if u r reading this post..kindly reply..) Another method: (Hope there is a better method than this too..) (100^3 + 101^3 +...+ 198^3) % 9 = ((198*199/2)^2  (99*100/2)^2) % 9 (using summation of cube series formula) = ((19701^2)(4950^2)) % 9 = (24651*14751) % 9 ............(a^2b^2=(a+b)(ab)) = 0.......... (Digital sum %9) Hope this helps.. Rohit 
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least val 11 and max val 7 
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Hi, some of the image files are missing. can u plz rectify it 
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hii sipra i'm new user of tg.actually its my first time to send any query to any one.so if i hav any mistake then plz forget him.can u help me actually my question on remainder.plz clarify method for solving to this question plz...ya i knw its soso question....but plz tell me its solution... 1.what is the remainder when 9^{1}+9^{2}+9^{3}+.........+9^{99} is divided by 6? _{83} 2.what is remainder when 63^{73} is divided by 13....? reply soon plz 
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hey tg group what happen guys if any body know about answers of these question then plz guy solve it and sent solution for my all query.plz guys 
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Hi Praveen, Is the answer 1. 3 2. 11 Kindly confirm, if correct i will post the solution. Hope this helps.. Rohit 
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Hii, Yes rohit u r correct. Answers are 3 & 11. OK help me out. In the above article in some of the solutions after finding out lambda( i dont know where is t8 symbol in compu and i found it just now !! ) he has divided it with 2. Check out the solution of 2^1990/1990 or the solution of 39^22/7 In both of them he has divided lambda by 2. Why ?? Thanx 
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Hi smruty, That is how this method works. Consider 39^{22} divided by 7 This is same as finding remainder when 4^{22} is divided by 7. Now 4 is 2^{2 }(a perfect square) and L(7) = 6 has a factor of 2. Hence we can divide L(7) by 2. Rule: Find the remainder when b^{k }is divided by N when N is relatively prime to b. If b is a perfect square and 2 is a factor of L(N) then find the remainder K^{'} when K is divided by L(N)/2 If b is a perfect cube and 3 is a factor of L(N) then find the remainder K^{'} when K is divided by L(N)/3 and so on and so forth for higher powers. For 2^{1990 }divided by 1990 2 and 1990 are not coprime. So Carmichael's Theorem cannot be applied. The method for tackling such problem is stated in the post itself by taking H.C.F and dividing N by that H.C.F. and solving further. Once you have divided 1990 (N in this case) by the H.C.F.(2,1990),there are no common factors between 2 and 995 (they are now coprime), hence we can apply Carmichael's Theorem. There is a typographical error in that problem 1990 is not 10 mod 995 but 1990 = 10 mod 396 Hope you have understood the correct version. If you have doubt in that reply back to this post. Hope this helps. Rohit p.s L(N) is lambda(N) 
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Very good article exhaustive and very informative. Clear all concepts though felt bit complicated at first but very useful. Thanks a lot for putting this articles. 
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Hi Guys, Sorry fosting post this question in this Forum. Can u solve this question? What is the product of all factors of the number N = 6^{4} x 10^{2}, which are divisible by 5? 12^{210} × 3^{102 }× 5^{140 } 22^{210 }× 3^{140} × 5^{105 } 32^{140} × 3^{210} × 5^{102 } 42^{140} × 3^{102} × 5^{210 } 52^{102 }× 3^{210} × 5^{140} Regards Pallav 
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Hi SE sir u had done a great job for beginers like me...... sir i had a boubt in t5he following question. find the remainder when _{33}34^{35 }is divided by 7.. remainder when _{50}56^{35 }is divisible by 11... thnx in advance sir. 
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Hi SE, Article is indeed great and will be very helpful in everyone's prepn. Thanks for posting it. Hi Praveen, There are lots of questions, similar to urs, already solved above in the article. Just go thru it and i m sure u will get it. Rgds Amar 
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HI SE SIR..CAN U PLS GIV D SOLUTIONS FOR THE FOLLOWING Q POSTED BY RAJA... find the remainder when _{33}34^{35 }is divided by 7.. remainder when _{50}56^{35 }is divisible by 11... 
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Hi Deepika, Please find the solution below: 33^34^35 / 7 33 and 7 are co prime hence lambeda= L(7)= 6 Now find the remainder of 34^35/6 using Carmichael # 1 theorem. p=HCF [34,6] = 2, q= 6/p =3 Hence remainder of 34^35/6 = remainder of 34^35/3 (apply Carmichael theorem) L(3) = 2 , 35/2 = 1 and it would give us 34^35/3 = 34/3 =1 and our expression becomes 33^34^35/7 = 33/7 = 5 remainder. Second question is also the same, try solving it or let me know if you need help. Thanks Amar 
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hi amar.... i have a doubt in basics itself.... how do v get 13^2=1 mod 24???? i hav tried d 2nd sum as per u said.... but got stuck at d end.... 50^56^35/11 50 and 11 are co prime hence lambeda= L(11)= 10 Now the remainder of 56^35/10 using Carmichael # 1 theorem. p=HCF [56,10] = 2, q=10/p =5 so d remainder of 56^35/10 = remainder of 56^35/5 L(5) = 4 , 35/4= 3 and it would give us 56^35/5 = 56/5 =1????? and then 50^56^35/11 = 50/11= 6 remainder????? reply fast 
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Hi Deepika, 1. 13^2= 169 when divided by 24 remainder is 1. 2. Which two questions you want me to explain?? 3. Where are u getting stuck, aa to gya answer 6 Rgds Amar 
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hi amar...thnx i thought that 6 is not d rite answer.... can u xplain me how d v get 5^10 = 1 mod 66 5^9*5=1 mod 66??????? few more quesd doubts n ques r from dis article itself) 1) 5^48 = 1 mod 153 5^46 * (35) = 1 mod 153???? 2) 39^22 divided by 7??? 3) 37^47^57 divided by 16???? dipika 
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Hello Dipika 1. 5^10 66= 11*3*2 hence L(66)= LCM [ L(11), L(3), L(2) ] =LCM [ 10, 2, 1 ] L(66) = 10 since b^L(n) gives 1 remainder hence 5^10 will 1 remainder. 2. I think u mean (5^9)*5 (5^9)*5 = 5^9 * 5^1 = 5^(9+1) = 5^10 since L(66) =10 hence 5^10 will give 1 remainder. 3. 5^48 = 1 mod 153 L(153) = LCM[L(17), L(9)] = LCM[16,6] = 48 hence 5^48 =1 4. 39^22 /7 L(7) = 6, 22/6 = 4 remainder 39^4/7 = (35+4)^4 /7 = 4^4 /7 = 16*16/7 = 4 remainder Don worry how others are doing it, make sure u r following the correct approach and getting the right answer. Rgds Amar 
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hi amar...thanx.... u r really helping me..... deepika 
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not a problem.. its the least we all can do for TG by helping each n everyone who comes at TG site and asks for help.. 
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hi amar...wats d appraoch 4 dis q... What is the product of all factors of the number N = 6^{4} x 10^{2}, which are divisible by 5? 12^{210} × 3^{102 }× 5^{140 } 22^{210 }× 3^{140} × 5^{105 } 32^{140} × 3^{210} × 5^{102 } 42^{140} × 3^{102} × 5^{210 } 52^{102 }× 3^{210} × 5^{140} deepika 
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1st of all I am really
sry Guys for posting TSD in this coloum reason being I hv posted twice
this prob but no 1 is checking TSD coloum. A starts from home for his office. He travels downhill, then on flat ground and then uphill to reach his office. It takes him 3 hrs to reach the office. On the way back home A takes 3 hrs 10 min to reach home along the same route. The speeds downhill is 60 km/hr, on flat ground is 48 km/hr and uphill is 40 km/hr. What is the distance between A’s home and his office? 144 km 148 km 154 km 160 km Data Insufficient Cheers !!!!!! Pallav 
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N = 6^{4} x 10^{2} = 2^{6 }*3^{4} * 5^{2} Method 1 : a number divisible by 5 will have 5 in it hence we need to check how many such numbers are possible. 2^0 * 3^0 * 5^1 2^0 * 3^0 * 5^2 2^0 * 3^1 * 5^1 2^0 * 3^1 * 5^2 2^0 * 3^2 * 5^1 2^0 * 3^2 * 5^2 . . 2^0 * 3^4 * 5^1 2^0 * 3^4 * 5^2 we recieved 10 such numbers for 2^0 now we need to do the same for 2^1 then 2^2..upto 2^6 which will give us total 70 such numbers. now to reach the answer we need to see what number is coming how many times in the product? if we multiply all the numbers having 2^0 power, scenario for 3 and 5 will be: 3^(0+0+1+1+2+2+...+4+4) = 3^20 similarly 5^(1+2+1+2...1+2) = 5^15 this is for first 10 numbers and when we take all the numbers (70) together, final power of 3 will be (3^20)^{7 }i.e. 3^140 and 5^(15)^{7} = 5^105 which gives us choice 2 as answer. Method 2: N = 6^{4} x 10^{2} = 2^{6 }*3^{4} * 5^{2} No. of divisors of N are 7*5*3 =105 Now out of these 105 numbers, 35 contain 5^0 power(these needs to be excluded as these are not divisible by 5) another 35 contain 5^1 and last 35 contain 5^2 power. if we multiply these 70 divisors and consider the scenario only for 5 : (5^1 * 5^1 * .. 35 times)*(5^2 * 5^2 * .. 35 times) from this product Final power of 5 will be 105 hence choice 2 as answer. Regards Amar 
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hey amar.... thnx a lot man deepika 
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Hi Pallav, I have replied to ur question in TSD section. Sorry for late reply man, i started my revision with number system topic and was checking posts related to it only. Gotta finish all within this week. Rgds Amar 
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hi amar... 1 more doubt 12^600 divided by 100??????? 12^20 mod 100 4^20 * 3^20 mod 100 76 * 01 mod 100???? i dint get dis step...can u help me out 
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Hi Deepika, 12 and 100 are not co prime you need to keep it in mind if you apply any remainder theorem here. However problem can be easily solved without getting into all that. 12^600 last two digits of the expression will be remainder when it is divided by 100. 12^600 = 4^600 * 3^600 = 2^1200 * 3^600 = (2^10)^{120} * (3^4)^{150} which is equal to 24^120 * 81^150 = 76 *01 = 76 answer because 24^even power = 76 and for a no. ending with 1, 10th digit = 8*0 =0 and unit digit is always =1 
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hey amar... thnx yar...i got d sum now... deepika 
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Amar your steps can be reduced as (10+2)^600 10^600 No remainder (2^10)^60 same 1024^60 LAST 2 DIGITS ARE 76 
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what is the remainder when sum of squares factorials of first 20 natural numbers is divided by 1152 ??? 
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What is the remainder of 34! divided by 71 ... anybody please help... 
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Superb article....!!!!!!!! 
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The One(Op), Was it *really* superb? ;) Was wondering, what things did you like and what you didn't? Would like your feedback/thoughts!  SE 
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sorry SE i dint understand anything from the beggining...frankly speaking i dint get how 13^2 is 1 mod24...i donno the concept of mod ...can u tell me page wer i can refer it 
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Prasanth, Conventions may vary from one person to another: 169 = 1 mod 24 169 mod 24 = 1 [169/24] = 1 169 = 24*7 + 1 but they all mean one thing: The remainder when 169 divided by 24 is 1. Mod = Modulo = Remainder produced by division operation.  SE 
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hey greata article. im confident of remainders now 
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hey i didn't understand nythng it ws all bouncer 
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what is the remainder when (17(9!) + 18!)
is divided by (9! * 8704 )???? Answer is 9! 
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dude d ans is 17(9!) & not 9! 
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Awesome and mind blowing article... It has been a privilege that we have such a teacher like u.. 
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Hi, I recently came across the que to find the remainder for (15^400)/1309 Can anyone pls let me know how to solve it. 
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IS the remainder 135.. 
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Hi Gaurav, Thanks for solving. But i recently came across this in one of the TIME tests and dont know the answer and how to solve. Can u pls elaborate on how u solved it 
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Hi Chetan 1309 = 7*11*17 So find the remainder of 15^{400} with 7, 11 and 17 individually and then combine them using Chinese Remainder Theorem to get the final answer. 
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Hi kamal, I know chinese theorem but will it not be two much of a time to solve it!!!! 
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15^{400} = 1 mod7 = 1 mod17 = 1 mod119 = 1 mod 11 = 1 mod1309. Was that lengthy? I didn't use paperpen to figure it out. 
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Using carmichael's theorem, someone solve this pl.. 3^4/6??? 
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2^1500 / 13 =???? 
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1 
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I'hv one question pertaining to remainder only, kindly throw some light. When 4^101 + 6^101 is divided by 25, the remainder is, options are {20,10,5,0}. 
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@amit 4^20/25 =1 (by Euler's theorem)so ((4^20)^5.4)/25=4 similarly ,for 6^101/25=6(remainder) so the ans is 10 
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I have a doubt ...in The number 84^86 when converted to base 210 ends in digit______??? +++++++++++++++++++++++++++++++++++++++++++++++++++ in one of ur discussion forum (http://totalgadha.com/mod/forum/discuss.php?d=5224), ther is a problem
in solution u have suggested to find out the remainder . the same concept u r using here(The number 84^86 when converted to base 210 ends in digit______????) to find out the unti digit ...Can u pls tell me hw is it possible that the same remainder will give u total no of trailing zeros as well as the unit digit. Help me.... 
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hi i have a doubt find the remainder when (123412341234....repeated 300 times) is divided by 909. i have solved this question and got my answer as 237. please check whether my answer is correct or not. thanks and regards nikita 
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question  we know that 2n! is divisible by (n! * n!) can u generalize its remainder? 
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hello sir. it's a faadu article . but one doubt i am able to solve problems v v v easily but plz give me full proof of carmicheal's theorem with an example. like 12^100 when divided by 67 gives what remainder. i know how to crack. jus solve 12^34 by 67 ;the same can be solved by binary system or any other method.but i didnet get in my mind that how we can say 12^34 or12^100 gives same remainder when divided by 67; or in other words i want to ask about detailed proof of carmicheal's theorem 
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Sir, Please help me with these ques: find remainder when (17)^36 + (19)^36 is divided by 11; 128^500 is divided by 153 
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1. 17^36 + 19^36 mod 11 Soln: 11 is prime. Thus, Euler function of 11 = 10 hence, 17^10 mod 11 = 1... similary 17^36= [(17^10)^3] * 17^6... First part would give a remainder of 1.. thus sum reduces to 17^6 mod 11... 17^2 = 289 leaves a remainder of 3 with 11. Thus find remainder of 3^3 with 11...which will give 5 Similarly solve for 19^36... will give a remainder of 3... Thus, finally add both to get the final answer as 8.. Please note that this is the basic approach to these kind of problems. 2. 128^500 mod 153 153= 17*9 Find individual remainders for 17 and 9 which would come out to be 16 and 4 respectively. Now the reqd. remainder R = 17x + 16 = 9y + 4.. Solving we get, x=3 and y=7.. substitute and you get the answer as 67... If someone has a better approach... please do mention. Regards, Amit 
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Hi Sir, As usual a great article again. Your level of work is really respectable. Sir if you can help me out with the following question, Ques: What will be the remainder when 128^1000 is divided by 153? Please help me out sir. 
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hi mohit, 92 will be the remainder ...use neg. rem here,u will get 25*82/153 then,simply multiply numerator terms and divide...will get 61/153 which is 92... 
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hi arsh, I too tried the same procedure but didnt get any of the options. can you explain ur method. And also the options given were, 1) 103 2) 145 3) 118 4) 52 
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hi mohit, i had checked it wid calci also 836(quotient) * 153=127908 which is 92 less than 128000...options are wrong man!!! i had used neg.remainder cncept..hope u know it!! 
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I guess, I've answered this question many times earlier also. Anyways, see it once more 153 = 9 × 17 So, we need to find the remainder obtained for the number 128^{1000} with the two divisors and then combine them using some method. Also there is no use of combining the two remainders obtained if options are available, as in this case. So here we just need to check that which option gives the same remainder with the divisors 9 and 17 as obtained by the number. Now let's find the remainder of the number with 9 and 17. 128^{1000} = 2^{1000} mod9 = (2³)^{333} × 2 mod9 = (1)^{333} × 2 mod9 = 2 mod9 = 7 mod9. So our answer should also be giving remainder 7 when divided by 9. Only 52 satisfies. So if answer is one of the options, then it should be 52. 128^{1000} = 9^{1000} mod17 = (9²)^{500} mod17 = (4)^{500} = 2^{1000} mod17 = (2^{4})^{250} mod17 = (1)^{250} mod17 = 1 mod17. That means number gives remainder 1 when divided by 17, so our answer should be like that also. And again 52 satisfies. Hence 52 is the answer. Kamal Lohia 
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thanks sir for correcting me!!!! 
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hello kamal sir, can u plz explain why am not getting the same answer by using neg.remainder concept???? 
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Arsh 25 is ok but what is this 82? 128^{1000} = (25)^{1000} mod153. And as HCF(128, 153) = 1 we can apply Euler's Theorem which says 128^{phi(153)} = 1 mod153. Now phi(153) = 153 (1  1/17)(1  1/3) = 96. So we have now 128^{1000} = (25)^{1000} mod153 = (25)^{1000} mod153 = (25^{96})^{10} ×25^{40} mod153 = 25^{40} mod153 = (25²)^{20} mod153 = 13^{20} mod153 = (13²)^{10} mod153 = 16^{10} mod153 = (16²)^{5} mod153 = (50)^{5} mod153 = (50)(50²)² mod153 = (50)(52)² mod153 = (50)(50) mod153 = 50² mod153 = 52 mod153. Kamal Lohia 
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hello sir, my mistake...from the starting ,am taking it as 128 * 1000 instead of 128^1000.. 
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hey mohit, yaar 52 will be the answer as explained by kamal sir, i had mistakenly took question as 128 * 1000 instead of the original statement as 128 ^ 1000. 
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hi kamal sir, Thanks for the ans. It is the right one. But one more thing, I'm not able to get how you get this 128^1000 = 2^1000 mod9 I guess it should be 2^7000 mod9. Please help me. 
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hi mohit, as 128/9= 2 thats why...i think u shud revise remainder theorem buddy!! 
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Thanks arsh... i got confused with other theorem... but yeah i will revise it again... 
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Guys hw to find last 2 non zero digits of any factorial??? take: 1) 154 ! 2) 200 ! 3) 89! plz xplain ???? 
Re: Remainders  
Hi viewpt kr I have already written much about this thing in my Math Corner. The first and last both posts talk about "How to find last one/two nonzero digits in a factorial number" mathematically. Hope it helps. Kamal Lohia 