How to find last two digits of a number  
This brilliant article was contributed to Total Gadha by Ashish Tyagi, a regular TGite. I do not think there can be easier or more lucid method of finding the last two digits of a number. Kudos to Ashish for this article. We will also welcome similar contributions from our other users. Do you have some useful or lifesaving funda that you would like to share? Send it to us and we will publish it with your name Total Gadha
I am dividing this method into four parts and we will discuss each part one by one: a. Last two digits of numbers which end in one Before we start, let me mention binomial theorem in brief as we will need it for our calculations.
Last two digits of numbers ending in 1 Let's start with an example. What are the last two digits of 31^{786}? Solution: 31^{786} = (30 + 1)^{786} = ^{786}C_{0 } ×— 1^{786} + ^{786}C_{1}_{ }×— 1^{785} ×— (30) + ^{786}C_{2} ×— 1^{784} ×— 30^{2} + ..., Note that all the terms after the second term will end in two or more zeroes. The first two terms are ^{786}C_{0 }×— 1^{786} and ^{786}C_{1}_{ }×— 1^{785} × (30). Now, the second term will end with one zero and the tens digit of the second term will be the product of 786 and 3 i.e. 8. Therefore, the last two digits of the second term will be 80. The last digit of the first term is 1. So the last two digits of 31^{786} are 81. Now, here is the shortcut: Find the last two digits of 41^{2789} In no time at all you can calculate the answer to be 61 (4 ×— 9 = 36). Therefore, 6 will be the tens digit and one will be the units digit) Find the last two digits of 71^{56747} Last two digits will be 91 (7 ×— 7 gives 9 and 1 as units digit) Now try to get the answer to this question within 10 s: Find the last two digits of 51^{456} ×— 61^{567} The last two digits of 51^{456} will be 01 and the last two digits of 61^{567} will be 21. Therefore, the last two digits of 51^{456} ×— 61^{567} will be the last two digits of 01 ×—21 = 21 Last two digits of numbers ending in 3, 7 or 9 Find the last two digits of 19^{266}. 19^{266} = (19^{2})^{133}. Now, 19^{2} ends in 61 (19^{2} = 361) therefore, we need to find the last two digits of (61)^{133}. Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 ×—3 = 18, so the tens digit will be 8 and last digit will be 1) Find the last two digits of 33^{288}. 33^{288} = (33^{4})^{72}. Now 33^{4} ends in 21 (33^{4} = 33^{2} ×—33^{2} = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21^{72}. By the previous method, the last two digits of 21^{72} = 41 (tens digit = 2 × 2 = 4, unit digit = 1) So here's the rule for finding the last two digits of numbers ending in 3, 7 and 9: Now try the method with a number ending in 7: Find the last two digits of 87^{474}. 87^{474} = 87^{472} ×—87^{2} = (87^{4})^{118} ×—87^{2} = (69 × 69)^{118} × 69 (The last two digits of 87^{2} are 69) = 61^{118} ×— 69 = 81 ×— 69 = 89 If you understood the method then try your hands on these questions: Find the last two digits of: 1. 27^{456} Last two digits of numbers ending in 2, 4, 6 or 8 There is only one even twodigit number which always ends in itself (last two digits)  76 i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as last two digits for even numbers. We know that 24^{2} ends in 76 and 2^{10} ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, 24^{34} will end in 76 and 24^{53} will end in 24. Let's apply this funda: Find the last two digits of 2^{543}. 2^{543} = (2^{10})^{54} ×—2^{3} = (24)^{54} (24 raised to an even power) ×—2^{3} = 76 × 8 = 08 (NOTE: Here if you need to multiply 76 with 2^{n}, then you can straightaway write the last two digits of 2^{n} because when 76 is multiplied with 2^{n} the last two digits remain the same as the last two digits of 2^{n}. Therefore, the last two digits of 76 × 2^{7} will be the last two digits of 2^{7} = 28) Same method we can use for any number which is of the form 2^{n}. Here is an example: Find the last two digits of 64^{236}. 64^{236} = (2^{6})^{236} = 2^{1416} = (2^{10})^{141} × 2^{6} = 24^{141} (24 raised to odd power) × 64 = 24 × 64 = 36 Now those numbers which are not in the form of 2n can be broken down into the form 2n ´ odd number. We can find the last two digits of both the parts separately. Here are some examples: Find the last two digits of 62^{586}. 62^{586} = (2 × 31)^{586} = 2^{586} × 3^{586} = (2^{10})^{58} ×— 2^{6} × 31^{586} = 76 × 64 × 81 = 84 Find the last two digits of 54^{380}. 54^{380} = (2 × 3^{3})^{380} = 2^{380} × 3^{1140} = (2^{10})^{38} × (3^{4})^{285} = 76 × 81^{285} = 76 × 01 = 76. Find the last two digits of 56^{283}. 56^{283} = (2^{3} × 7)^{283} = 2^{849} × 7^{283} = (2^{10})^{84} × 2^{9} × (7^{4})^{70} × 7^{3} = 76 × 12 × (01)^{70} ×—43 = 16 Find the last two digits of 78^{379}. 78^{379} = (2 × 39)^{379} = 2^{379} × 39^{379} = (2^{10})^{37} × 2^{9} × (39^{2})^{189} × 39 = 24 × 12 × 81 × 39 = 92 Now try to find the last two digits of Ashish Tyagi I am afraid 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: How to find last two digits of a number  
SEEMS AWESOME....... 
Re: How to find last two digits of a number  
stupendous man !! gr8 wrk.. 
Re: How to find last two digits of a number  
hi if problems last digit is 5 like 25^{45 } ^{den wat will be last digits} 
Re: How to find last two digits of a number  
gr8..... 
Re: How to find last two digits of a number  
Thanks a lot sir it's awesome. Just wanted to know for No. ending in 5, the last two digits would be 25 always? 
Re: How to find last two digits of a number  
Hi Sir, I have a doubt i want to find the last 2 digits of 38 to the power 221? Thanks 
Re: How to find last two di*gits of a number  
Hi can any one explain me how to calculate last 2 digit of 2^2^2008. 
Re: How to find last two digits of a number  
2^2^2008? = 4^2008 =2^4016 =(2^10)^401 * 2^6 =(1024)^401 * 64 we know that 24 raised to odd power ends with 24 therefore 24 * 64 the ans should be. =36 
Re: How to find last two digits of a number  
thanxs for the correction 
Re: How to find last two digits of a number  
very very nice sir..thanku v much!!! 
Re: How to find last two digits of a number  
Very nice topic Ashish...Thank you very much 
Re: How to find last two digits of a number  
thanks Ashish and TG... its really great method. 
Re: How to find last two digits of a number  
Thanks Ashish for such useful fundas.. 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
nice article... 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
last two digits can also be find out using the cyclicity for ten's place. digits cyclicity 2,3,8 20 4,9 10 5 1 6 5 7 4 
Re: How to find last two digits of a number  
Could anybody help with problems of the type 2007^2008^2009 please??? How do we go about this kind of problems? 
Re: How to find last two digits of a number  
Hello Ruchi The ans will be 1. 61 2. 39 3. 61 Regards Jinson 
Re: How to find last two digits of a number  
hi..jinson .. i second u wh the below answers:: 1. 61 2. 39 3. 61 Best Regards Randeep 
finding last two digits of a number ending wd 5..  
finding last two digits of a number ending wd 5.. (.......E5)^{even }= 25, where E represents even no. 75, otherwise 
Re: How to find last two digits of a number  
kudos to u ashish.. that was brilliant.. thanx a lot... 
Re: How to find last two digits of a number  
answer 2 d above questions related 2d problems asked in finding d last 2 digits of nos ending wt 2,4,6,8 r... 1) 16 2)12 r dey correct TG sir???? 
Re: How to find last two di*gits of a number  
the only problem with 76*2^n rule is when 76*2 gives 152 last two digit is 52 not 02 so we all can note this exception...... 
Re: How to find last two digits of a number  
thnx for all above concepts...dey will be of great help.. 
Re: How to find last two digits of a number  
thx ashish 4 dis splendid job.....sir u dnt know how much great work u have done i dnt have any words in mah lexicon 2 describe it simply hats off 2 u... 
Re: How to find last two digits of a number  
really useful man.. thnks a lot 
Re: How to find last two digits of a number  
i didnt get the soln.. can u help me out with it 
Re: How to find last two digits of a number  
the most lucid concept i hv evr studied..thanks tg sir 
Re: How to find last two digits of a number  
Dear Ashish, Thank you very much for this valuable post. Regards. 
Re: How to find last two digits of a number  
Thanks a lot TG...really a very helpful article...nd thanks to ashish as well.. Regards 
Re: How to find last two digits of a number  
Great concepts 
Re: How to find last two digits of a number  
awesome article ...i havnt seen calc easier than this one ....nice .... 
Re: finding last two digits of a number ending wd 5..  
dis method z nt valid..plz check it agn...mc 
Re: How to find last two digits of a number  
Hello
These methods are just awsome. time saving, fast and verbal calculation. Great Job Done. 
Re: How to find last two digits of a number  
Dude i m getting the same answers. ok tell me 1 thng hw much time u took to solve these 3 ques.? 
Re: How to find last two digits of a number  
Hello Sahana I have got 01 as the answer of this Problem. Is this Right ? 
Re: How to find last two digits of a number  
hey pallav .. can you please explain how you arrived at the answer?? 
Re: How to find last two digits of a number  
awefome 
Re: How to find last two digits of a number  
What if we cud have 2003^2004^2005 and so on.... and need to find last 2 digits..... 
Re: How to find last two digits of a number  
the last 2 digits of 2003^2004^2005 are 81 
Re: How to find last two digits of a number  
Hi can anyone please tell me the detailed Solution for this Question. 2003^2004^2005..... LAST 2 DIGITS 
Re: How to find last two digits of a number  
Hi Pallav!! the Ans m gettin for 2003^2004^2005..... is 81. Plz tell me d correct ans... 
Re: How to find last two digits of a number  
nice one indeed very useful.... thanx 
Re: How to find last two digits of a number  
2003^2004^2005 L(100)=20 2004^2005 mod 20 p=4 and q=5 2004^2005 mod 5 (1)^2005 mod 5 1 mod 5 R=4x=5y1 x=1, R=4 2003^4 mod 100 81 mod 100  SE 
Re: How to find last two digits of a number  
i second d above solution..that is the right way to do.. 
Re: How to find last two digits of a number  
I think that will fail in case of 2222^4000^6666. Last two digit of each term, 22^00^66 = 22^0 = 1 but, 2222^4000^6666 = 76 mod 100. 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Hi guys/Gowtham Muthukkumaran Thirunavukkarasu In this forum one of the reply posted by Gowtham Muthukkumaran Thirunavukkarasu Posted on 15 May 2009. He (u) said that product of any numbers ending with 5 always ends in 25. I think it is not necessary as it could be either 25 or 75. Take for example: 65 x 35 = 2275 85 x 35 = 2975 or many more ... which in not 25 but 75. Tell me if 'am wrong or not getting what he wants to say. 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Hi Raju, Good observation.. He made a mistake.. Anyways, the product of any two numbers ending with 05 always ends in 25 Rohit 
Re: How to find last two digits of a number  
Yup Pallav! u r right!! Dat wz indeed a short method... Thnx alot 
Re: How to find last two digits of a number  
can u explain it in detail!!!! 
Re: How to find last two digits of a number  
Hi Guys, can anyone help me wid the detailed solution of this question. 6^83 + 8^83 DIVIDED BY 49 WHAT IS THE REMAINDER ? Pallav 
Re: How to find last two digits of a number  
hi, how number 2^xxxx56 will translate into (2^10)^odd number)*2^6 ?? 
Re: How to find last two digits of a number  
Hi Rohit Thanx for your reply but i cud'nt get your explanation. Can you please explain me in Detail. Thanks 
Re: How to find last two digits of a number  
its great...... 
Re: How to find last two digits of a number  
amazing ! simply amazingggggggggggggg 
Re: How to find last two digits of a number  
simply awesome and amazing!!!!!! 
Re: How to find last two digits of a number  
ultimate!! cheers!! 
Re: How to find last two digits of a number  
Aastha, rather than applying Euler's Theorem, use the methods described in Remainders. Amandeep, please refer this. 
Re: How to find last two digits of a number  
Hi SE, thanks for replying as i have been waiting for it for long now! i was trying to avoid carmichael's thm and my effort was to solve all remainder question by EUlers thm. but as u suggested i studied the Carmichaels's thm,,, as i found it easier.. actually i observed,, it reduces the totient further, lesser that euler totient and as we reduce the totient, it leaves lower value of when divided by the power hence earier calc in finding remainder of a number lamda(16) is 4 and phi(16) is 8... hope i made sense.. thanks again... 
Re: How to find last two digits of a number  
awesome..... 
Re: How to find last two digits of a number  
Great article...thank u!!!! 
Re: How to find last two digits of a number  
gr8 article thankzzz 
Re: How to find last two digits of a number  
@life goes on .. awesome awesome awesome.. wat i was doing was very lengthy.. 
Re: How to find last two digits of a number  
GR8 wrk.... i admire u 4 such wondrful work....... kudos 2 u.... keep it up.... 
Re: How to find last two digits of a number  
Dear CAT Aspirants, I would advise you all to practice these short cuts thoroughly before you go to take on CAT. I read this topic and remainders topic yesterday. Today i gave CL mock test and i was so happy to see a question on last two digits of a number. But trust me, i could not remember what was the short cut for solving the question. Wasted 2 minutes on this but all i could remember was '24' , 'euler theorm' and 'lambeda' . Switched back to basics and problem got solved. Thank GOD fetched me 3 marks.. So in nutshell ..practice and practice more.. ATB Below is problem which came in test: find the 10^{th} digit of the number(i.e. 2nd from right) : _{6}^{117} 
Re: How to find last two digits of a number  
I also went through same. Became desperate. Wasted 56 min and ultimately bowed out. 
Re: How to find last two digits of a number  
Dont worry Smruty. Its good if we do all our mistakes in mocks and take a lesson from each of them. Ultimately what matters is our performance in CAT paper. As per your reply it seems you had left the question, i would recommend you to go through remainders post. It would give you better insight to solve such questions.. 
Re: How to find last two digits of a number  
Find the last two digits of: 2. 79^{83} 3. 583^{512} i am getting 1. 61 2. 39 3. 61 can u pls xplain me how did u get 2nd n 3rd answers 39 n 61...i am getting 79 n 41.... pls reply soon 
Re: How to find last two digits of a number  
Hi Deepika, Although the question was not asked to me but i hope u wont mind if i answer this. It goes like this: 79^{83}= (79^{2})^{41} * 79 = xx41^{41} *79 Now you need to refer to the rule given above to find the last two digits of a number ending with 1. From this rule : 4*1 =4 = 10th digit and unit digit =1 hence our expression becomes 41*79 = xx39 which gives 39 as last two digits Now try to solve the 3rd question following the same rule and m sure u will crack it Rgds Amar 
Re: How to find last two digits of a number  
hey thnx amar..... i got d answer...i did d last step rong... thnx 1 again.... n total gadha,,,,u rock man........... 
Re: How to find last two digits of a number  
What if we cud have 2003^2004^2005 and so on.... and need to find last 2 digits..... ans: firstly consider 2005^2006^2007.....Here the last 2 digits ll cum out to b 05...Since 5^(any no.) is alwyz 25 Now consider (2004)^xxxx25 it can b written as (4*501)^xxxx25 = 4^xxxx25 * (501)^xxxx25 =2^xxxx50 * 01 = (2^10)^xxx5 * 01 =(1024)^xxxx5 * 01 = 24 * 01 = 24 Finally the question is reduced to 2003^xxx24 which is of the form = (03 * 03 * 03 * 03)^ xxx1 or 03 * 03 * 03 * 03)^ xxx6 = (81)^ xxx1 or (81)^xxx6 In both of the cases last 2 digits cum out to b 81..... can u pls simplify it...i dint get it 
Re: How to find last two digits of a number  
it means that only last two digits of 2003 will contribute to last two digit of the final product. Hence 2003^{xxx24} can be written as 03^{xxx24} which in turn can be written as (03^{4})^{xxx1} or (03^{4})^{xxx6 because any number xxxx24 when divided by 4 will give 1 or 6 as unit digit of its quotient.Hope this helps.} 
Re: How to find last two digits of a number  
Hi Amar I am getting 56 as the last 2 digit of your mock question. 6^11^7 Am I Right? 
Re: How to find last two digits of a number  
Absolutely correct Pallav 
Re: How to find last two digits of a number  
Hi Amar you are enrolled in which center of CL. & how was the Mock 8 level this time. I am also a CL student but I didnt take Mock 8. Cheers !!!!! 
Re: How to find last two digits of a number  
I am enrolled with Noida centre, test series student. Proc mock 8 was easy as compared to previous tests. 22 questions in each section with +3 /1 marking scheme. DI was the easiest, QA average and EU on the tougher side(coz of tricky questions). Hope we are not violating terms and conditions of use of Totalgadha.com by discussing a competitor's (CL) stuff here. However if i did, i am very sorry and i simply dint mean to do so. TG is the best and everyone knows this, one who doesn't has not visited the site properly. 
Re: How to find last two digits of a number  
My Dear Last 2 digit for any power of 5 will always 25.
Try it... 
Re: How to find last two digits of a number  
Anand I dont know who is 'dear' here but i can give u an example of what ppl are talking : 5^3 = 125  i 15^3 = 3375  (ii) (ii) implies that if 10th digit is odd(1 here) and power(3 here) is odd then last two digits are not 25. 
Re: How to find last two digits of a number  
Hi Deepika and Netra, In regard of question 2003^2004^2005 I am not able to get the first line of both solutions. That is we have to find tens and unit digits of 2004^xxxx25 . From where this 25 had come when it was initially 2005 Thanks In Advance .. 
Re: How to find last two digits of a number  
uhhh!! Last 2 digit for any power of 5 5^1 =05 5^2 =25 5^9 =1953125 5^11 =48828125 5^19 =.......
not multiple of 5..... 
Re: How to find last two digits of a number  
Bang on the target!!!!!!!!!! Must say Fundooooooooo article TG....... U rock!!!!! 
Re: How to find last two digits of a number  
hey.......please tell me how did you get that.... last digit 56..... please elaborate it...... 
Re: How to find last two digits of a number  
versha jhunjhunwala 6^11^7 (2^11^7)*(3^11^7) (2^11^7) ok as per TG's Concept.. 11^7 last 2 digits wud be 71 nw 2^71 which can be written as (2^10)^7 * 2^1 1024^7*2^1 = 24*2 as the last 2 digits.... 48 same can be done with 3 3^11^7 3^71 (3^4)^17 * 3^3 81^17 * 27 61*27 last 2 digits are 47 nw multiply 48*47 So we get 56 as the last 2 digits of the ques. Hope its understandable Cheers !!!!! Pallav Jain 
Re: How to find last two digits of a number  
Thanks ...Pallav.... I understood it .... but is it necessary to frst calculate lke u did ...11^7 n den fnd it 6^that power ...
I mean if i went lke dat .. [{6^11}^7] so ...6^11 n cyclicity of 6 in case of ten's digit is 5 ..so [{6^5*2 }*6^1]^7 [___56]^7 n we will get ... ___16 so d ans wud be ...16 (is it wrong ...?)
2. 2003^2004^2005 in dat case: [{2003^4}^501]^2005 [{___81}^501]^2005 [__81]^2005 41 wud be d ans... But ans is 81 ...so my approach is wrong or wat??

Re: How to find last two digits of a number  
Sry ...the ans shud be .._01 according to me... 
Re: How to find last two digits of a number  
i m really sorry varsha i didnt get ur explanation. 
Re: How to find last two digits of a number  

Re: How to find last two digits of a number  
Hi Kanika Is the answer 64?? 
Re: How to find last two digits of a number  
hii ol What will be the last digit of ? can u solve dis 4 me!! kanika Answer is 6 
Re: How to find last two digits of a number  
i did it again..again gt ans as 4.. can u solve it here pallav?? 
Re: How to find last two digits of a number  
answer s 4 evn i got 6 !! datz y m askin plz teme tapas plzz 
Re: How to find last two digits of a number  
Pardon me cz I hav neva posted anywhere..so dunno how to make special mathematical symbols..hehehe..but i hope this is understandable... start from the first term: in the first term 4^{5 }= 1024. next 3^{1024 }= [81]^{256}. Now last two digits of this number (with the method given by TG sir) = 81 So, the first term reduces to 2^{81} . Writing it as [[2]^{10} ]^{8} * 2 = [24]^{even power} * 2 = 76 * 2 = 52 Now take the second term. Start from the top: 3^{5} = 243. Next we have 15^{243} = 3^{243} * 5^{243} = [81]^{60} * 3^{3} * 5^{243} Now this reduces to (01) * 27 * (25) { as 05 raised to any power always ends in 25} this again reduces to (27) * (25) = (75) Now the final 3^{(75)} = [81]^{(18)} * 3^{3} = (41) * (27) = (07) So, finally multiplying the two terms (52) * (07) we have the units digit as 4 and tens digit as 6 
Re: How to find last two digits of a number  
thanku thanku thanku 
Re: How to find last two digits of a number  
hi...it can be solved using the reminder theory also R(2^3^4^5/10)... let's move from the bottom... 1) we need 3^4^5 in the form 4K+reminder, (since 2^4k divided by 10 gives reminder 1) 2) to write 3^4^5 in the form 4K+reminder, we need to write 4^5 in the form 2k+reminder (since 3^2k divided by 4 gives reminder 1) 3) but 4^5 is always of the form 2k 4) so, 3^4^5 becomes 3^2k.. 5) when 3^2k is divided by 4 (as required by step 2)the reminder is 1...so 3^2k can be written as 4K+1 6) so 2^3^4^5 can be written as 2^(4K+1) and when divided by 10, the reminder is 2 (because 2^4k will give reminder 1)...so the last digit of 2^3^4^5 is 2... similarly the last digit of the next term is 7.. so the product is 14 and hence the last digit is 4!!! this method can be used to find the last digit of a number containing any number of powers... hope, i didn't confuse anyone 
Re: How to find last two digits of a number  
bt frst method s far betr 
Re: How to find last two digits of a number  
@ Kanika n Tapas, use this shortcut method in the second case of 3^15^3^5. as 3^5 = 243 therefore, 15^243= xx75 (as last two digits). How? see If number ends in 5 then use this format (X5)^n for n>1 1) if X is even then for all value of n the last two digits always be 25. 2) if X is odd then two case: a) if n is even then also last two digits will be 25. b) if n is odd then last two digits will be 75. This is alway applicable if number ends in 5 So in case 15^243 as X is odd and so is n hence last two digit is 75. There I have seen number of people say that (XX5)^n always ends in 25! no some time it may be 75. If single digit 5^n then it always end in 25 I think kanika had done same mistake thats why get 6 as answer. hope it helps u someway. Thanks 
Re: How to find last two digits of a number  
Hey TG...Simply fantastic... My ans

Re: How to find last two digits of a number  
Awesome article Mr. Ashish [though I know i am little late in admiring this].... Thank you 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
2^1= 02 2^20=1048576 Therefore 2^21= 2097152 therefore the cycle for 2 cannot be 20 For 3 & 8 it's 20. because 3^20= .......01 & 8^20= ........76 for 6 also I find the same thing 6^1=06 6^5=7776 6^6=46656 So the cycle is not repeating after 6^5 
Re: How to find last two digits of a number  
hi if problems last digit is 5 like 25^45 then d last two digit answer will be 25 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
awesome.. wow how did u find it its truly helpful.... n saves alot of time too 
Re: How to find last two digits of a number  
superb.... 
Re: How to find last two digits of a number  
3^4^5 last two digits is 01. 3^4 is 81 81^5 last two digits is 01 
Re: How to find last two digits of a number  
Hi All, Could you please let me know how to solve below questions?
1.) (12^55/3^11)+(8^48/16^18) will give the digit at unit place as Ans  (a)
Ans  (c) 3.) The power of 45 that will exactly divide 123! is Ans  (c)  But here i am getting answer (a) 4.) Three numbers are such that the second is as much as lesser than the third Ans  (e) Thanks Rahul 
Re: How to find last two digits of a number  
Hi Rahul 1) (12^{55}/3^{11}) + (8^{48}/16^{18}) = (3^{44})(4^{55}) + 2^{72} = (a1)(b4) + c6 = d0. (d) 2) [x]^{2} + {x} = 25.16 = 25 + 0.16 = (5)^{2} + 0.16 = (5)^{2} + 0.16 So x = [x] + {x} = 5 + 0.16 = 5.16 or 5 + 0.16 = 4.84. (c) 3) 45 = 3^{2}*5 and 123! contains 3^{41+13+4+1} = 3^{59} or 9^{29} and 5^{24+4} = 5^{28} so 45^{28}. (a) 4) Let the three numbers be a, b, c and given is: c  b = b  a also ab = 85 and bc = 115. So ab + bc = b(a + c) = b(2b) = 2b^{2} = 85 + 115 = 200 = 2*10^{2}. So b = 10 (e) 
Re: How to find last two digits of a number  
Hi kamal, Thank you for your quick response.It seems easy now the way you explained it. For q. 1) I also got answer as (d) but book (Arun sharma's) says it as (a) For q. 3) I also got answer as 28 but book says 31. Rest are correct and I got it.
Thank you once again

Re: How to find last two digits of a number  
for 2nd ques 4.84 is not possible because [x]=5 and {x}=48 then [x]^2+{x}=25+.48=25.48 option a is correct. 
Re: How to find last two digits of a number  
ans is 4 ..... 
Re: How to find last two digits of a number  
ans will be 01. 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Hi Gowtham Muthukkumaran, In your post you said "The product of any 2 numbers ending with 5 always ends with 25" but see this: 15x25=375 25*25=625 25*35=875 25*45=1125 ... and so on I think it should be " The product of any 2 numbers ending with 5 always ends with 25 if sum of their digits in 10s place is even and it ends with 75 if sum of their digits in 10s place is odd." this is for 2 digit numbers. 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Find the last two digits of: 2. 79^83 answer i am getting is 79^2^41*79 (79*79=6241) so 41^41*79 then 4*1 =4 answer is 49 i don know i am correct or wrong please reply tg 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Hi Bhaskar 79^{83} = (79)(79)^{82} ≡ (79)(21)^{82} mod100 ≡ (79)(41) mod100 ≡ 39 mod100. Kamal Lohia 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
thanks tg team 
How to find last two digits of a number  
hii everyone.. I am new to this site and its awesome for we CAT aspirants. Intially I was using the remainder theorem method to find out the last two digits. But after knowing these methods/shortcuts..its further reducing the time and effort to calculate last two digits. what i think using both therse methods simultaneously one can solve these type of questions in a very short time. 
Re: How to find last two digits of a number  
Finding the last two digit makes sense when one understands the concept underneath n the number pattern it follows.... It can give some edge over No. Pattern as the CAT itself lies in NUMBERS only. 
Re: How to find last two digits of a number  
Awesome article as well as Comments. Both helped me a lot. Cheers 
Re: How to find last two digits of a number  
superbbbb

Re: How to find last two digits of a number NUMBER TABLE Gowtham  
To TG Sir, I am from Mumbai intending to join the CAT CBT club & have sent a cheque to that effect through speed post 3 days ago.I have no idea whether it reached TG or not.How do I know? 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
please mail the following details at admin@totalgadha.com 1. cheque no 2. correspondence address at which you sent the post. 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Sir i mailed as u asked me to.Still no clue whether the cheque reached u or not. 
Re: finding last two digits of a number ending wd 5..  
How it works yar..... (....O5)^Odd = 75 (...E5)^Odd/Even =25 (...O5)^Even = 25 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Just take the sum of tens place.. even gives 25 odd 75.. more generalized.. 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
hi everyone.............. how to find last two digits of multiplication 1089 * 1089?? any shortcuts........ 
Re: How to find last two digits of a number  
Let a = 1! + 2! + 3! + . . . . .. . .. 99! + 100! Number of digits in the number ‘a’ are How to do this problem? 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Multiply the last 2 digits of the numbers ie, 89*89 = 7921 so last 2 digits of1089*1089 is 21.

Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Hi Lakshmi It becomes easier if you understand that finding last two digits of a number is same as finding remainder of the number with 100. So in this case we have: 1089×1089 ≡ (89×89) mod100 ≡ (11)(11) mod100 ≡ 21 mod100. So required last two digits are 21. Kamal Lohia 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
why 26^n always ends with 76 if n>1....sir please reply me on merodreamisland@gmail.com 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Hi manish Look carefully. It says 24ⁿ ends in 76 as last two digits if n is even otherwise it ends in 24 only. (Here n is a natural number). Reason is that 76 is a special number (2digit automorphic number), which when raised to any natural number power ends in same last two digits i.e. 76 always. (76ⁿ ≡ 76 mod100). Also, there is one more speciality with 76 as follows: 76 × 2ⁿ ≡ 2ⁿ mod100 (iff n > 1). Kamal Lohia

Re: finding last two digits of a number ending wd 5..  
hi yogesh... its always 25..... n nt 75. 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Let a = 1! + 2! + 3! + . . . . .. . .. 99! + 100! Number of digits in the number ‘a’ are a.137 b.283 c.314 d.189 e.none of these How to do this problem? 
Re: How to find last two digits of a number  
Last digit of any number ending with 5 will be 5 itself 
Re: How to find last two digits of a number  
just awesome...article 
Re: How to find last two di*gits of a number  
1+4+6+5+11+6+... 200terms =>(1+6+11+16+...100terms)+(4+5+6+7+...100terms) =30200 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  
Hi ,this is really helpful. thanks Gowtham. 
Re: How to find last two digits of a number  
Hi Sir, How to find last two digits of 14^14^14?? Thanks, neha 
Re: How to find last two digits of a number  
Hi Neha Finding last two digits of a number is same as finding its remainder with 100 i.e. 4 and 25. Now given number is divisible by 4, so we just need to find its remainder with 25 and then combine. We know that 14 and 25 are coprime, so 14^{20} = 1 mod 25. Also 14^{14} = 20k + 16 => 14^{14^14} =14^{16} mod25 = 4^{8} mod25 = 2^{16} mod25 = (2^{6}) mod25 = 64 mod25 = 11 mod25 = 36 mod25. As number is divisible by 4, its last two digits are 36. 
Re: How to find last two digits of a number  
Hi Sir, Thanks for the solution.. Just want to ask though 11 is positive remainder with 25 u hv added 25 to it since obviously for 14 number should end with 6??? And another question: Whats the last two digit for: 11^11!+12^12!+13^13!+.... + 80^80! Regards, Neha 
Re: How to find last two digits of a number  
@neha as in 11^11!+12^12!+.........+80^80! 11! = 11*10*9*.........*1 it will form 10*5*4*2=400 means power of each term will b multiples of hundred so 11^11! will end in no.of zeroes. hence last two digits of above sum wil be zero so Ans.= 00 
Re: How to find last two digits of a number  
For: last two digits of 14^14^14?? > 14^14^14 = 14^(2*7)^14 consider (2*7)^14 => 2^14 * 7^14 => 2^10 * 2^4 * (7^4)^3 * 7^2 => __24 * 16 * __01^3 * 49 => __16 Therefore 14^14^14 => 14^__16 => 2^__16 * 7^__16 => (2^10)^odd no * 2^6 * (7^4)^4k => __76 * 64 * (__01)^4k => 64
why am i getting 64? Is it correct? 
Re: How to find last two digits of a number  
Hi Rhythm Goyal Everything was fine except secondlast step where it should be 24 in place of 76 and the answer will be 36. 
Re: How to find last two digits of a number  
Oh... Thats so stupid of me... I have verified my calcluations thrice before posting. Still got same wrong answer all the 3 times. Thanks a lot Sir. I may now not comitt such in CAT. Sir, I need to discuss regarding Regular batch / CBT. Where can we discuss? 
Re: How to find last two digits of a number  
Any power of five (>1) always has 25 as last 2 digits 
Re: How to find last two digits of a number  
(2957^3661)=???? 
Re: How to find last two digits of a number  
(2957^3661)=57 
Re: How to find last two digits of a number  
awesome work by sir but sad to see that more knowledge given to premium students. 
Re: How to find last two digits of a number  
Hi, I used this approach. the no is 9483^67483 (83^4)1687*83^3 Now, (83^4)^1687 results in 41 as the last two digits.
83^3 results in 87 we multiply 87 and 41 and we get 67..?Bt dats incorrect. Plz explain?? 
Re: How to find last two di*gits of a number  
i calculated this way : (38)^{221} =(19*2)^{221} =19^{221} *2^{221}=(19)*(19^{220 })*(2)*(2^{220 }) =19*(19^{2})^{110} * 2(2^{10})^{22} =19*(361)^{110} *2(24)^{22} =19*(last digit is 61)^{110} *2( last digit is 76) =19*(61)^{110} *2(76) =19*(01)*2(76) now if we multiply:19*01*02*76=last is comming 88 Ok, but as u have mentioned 2^{n} *76= last two no. of 2^{n} ,therefore 19*01*2^{1} =last digit is 44. how ... ^{} 
Re: How to find last two digits of a number especially for power of 2  
ya i too faced the plb ....when cal. =19*01*2*76,it should 19*02, but the ans is not comming 
Re: How to find last two digits of a number  
u said 2^3^2 is ~ 2^9 but a^b^c = a^bc So 2^3^2 should be 2^6 so 2^2^2008 = 2^4016 (& we should go directly for this)

Re: How to find last two digits of a number  
simply great..... 
Re: How to find last two digits of a number  
The mistake which you are doing is dat u are starting from the left most side...u shud try solving from the right most side or the uppermost power. 
Re: How to find last two digits of a number  
some one please tell me how to solve the following question. q. find the last the last 2 digits of 62^43^54^65^76^87. 
Re: How to find last two digits of a number  
hey arun can u please enlighten me on the 4th line of ur answer ("the 1)") how the remainder will be 1 when 2^4k is divided by 10???? 
Re: How to find last two digits of a number NUMBER TABLE Gowtham  

Re: How to find last two digits of a number  
you r correct jinson 
Re: How to find last two digits of a number  
I am unable to find(23)^8 forget 3,7,8 raised to some powers... Do I have to by heart the values for such things ? I'm sure to goof up if this continues ! Please help ! var __chd__ = {'aid':11079,'chaid':'www_objectify_ca'};(function() { var c = document.createElement('script'); c.type = 'text/javascript'; c.async = true;c.src = ( 'https:' == document.location.protocol ? 'https://z': 'http://p') + '.chango.com/static/c.js'; var s = document.getElementsByTagName('script')[0];s.parentNode.insertBefore(c, s);})(); 
Re: How to find last two digits of a number  
5 there are two case one (oddno5)^odd= 75 in all other cases it is 25.. oddno represent tens digit 
Re: How to find last two digits of a number  
one of the better time management tips expereinced... thnq for the guidance... 
Re: How to find last two digits of a number  
Hi, We have 6 numbers here. 62,43,54,65,76,87. See since we have one 5 only in (65) there is no other 5 in any number.And we have plenty of 2s ( in 62,54,76) so we will get 0 in the units digits. So now once we have found the units digit we need to take care of the tens digit. Now remove the 5 and 2 which we have found so you are left with 31,43,54,13,76,87. Find the last digit of the product of above numbers it is 2. So the answer is 20. 
Re: How to find last two digits of a number  
so 11^11! will end in no.of zeroes..>>>>.> this logic is not correct.it is wroong 11^ (.....00) will still have 1 in units digit. 
Re: How to find last two digits of a number  
Hi Neha/Vinayak see my answer below and tell me if anything is wrong. consider 11(11!) we will find the last 2 digits of 11(11!). let n=11! so we have to find the binonial expansion of (10+1)^n. the last two terms will be NC(N1) *10*1 + 1*1*1 =N*10*1 =11!*10*1 SO IF we add for all the terms we get (11!+12!+80!)*10 + (1+1+1 70 times) so the last 2 digits will be 70. IF someone finds my answer wrong please tell me. 
Re: How to find last two digits of a number  
Hello friends, How are you getting 61 for the 1st question? Isn't it 41?? Please reply... 
Re: How to find last two digits of a number  
Hi Dhaval , Please see below 27^456 (27^2 * 27^2)^114 now we find the last 2 digits of 27^2 they are 29 so we have ((..29)(..29)) ^ 114 the last 2 digits of (..29)(..29) will be (..41) so finally we have the number (..41)^ 114 the the units digit will be 1 tens digit will be units digit of 4*4=16 so tens digit is 6 so answer is 61 
Re: How to find last two digits of a number  
There is a rule, if either of these 0,5,6 is a last digit then its always remain the same (i.e, 0,5,6 resp;) eg: 25^n: last digit remains 5 similarly with 0 and 6 
Re: How to find last two digits of a number  
Hello sir.!What are the last two digits of 136^132! ?by using above method.. 
Re: How to find last two digits of a number  

Re: How to find last two digits of a number  
52 
Re: How to find last two digits of a number  
soweii bro bt i didnt understood wat u mean by saying divide by 100 ?? plzzz help me understand plz 
Re: How to find last two digits of a number  
Please explain this logic?

Re: How to find last two digits of a number especially for power of 2  
2^10^(even no)will always have a remainder 76 and 2^10^(odd number)will always have a remainder 24 
Re: How to find last two digits of a number  
76 ? 
Re: How to find last two digits of a number  
Hey.........i agree with you that division by 100 is an available method for finding the last two digits......but the method given above gives us a shortcut.......thereby saving time.....and thanks for the concept of finding the last two digits for nos ending in 5............ 
Re: How to find last two di*gits of a number  
sum of this series should be 20200, (4+5+6+7+...103)+(1+6+11+16+21+...496), sum of arithmetic progression is n/2(1st term+last term). 
Re: How to find last two digits of a number  
Thanks a lot !! This is the by far the best thread I have read till now which explains it in such an elegant details !! Many thanks to the author. 
Re: How to find last two digits of a number  
Find the last two digits of 3^(33)^(333)? 
Re: How to find last two digits of a number  
Hi Abheek 3^{20} ends in 01 as the last 2digits and when we raise 01 to any power, last digit will remain same as 01 only. So first we need to find remainder of the power i.e. 33^{333} with 20. Now 33^{333} = (7)^{333} mod 20 = (7^{333}) mod 20 = (7^{4})^{83}(7) mod 20 = 7 mod 20 = 13 mod 20. So we simply need to find the last two digits of 3^{20k+13} i.e. 3^{13} which are same as that of (3^{4})(3^{4})(3^{4})(3) i.e. (81)(81)(81)(3) i.e. (61)(81)(3) i.e. (41)(3) i.e. 23. Kamal Lohia 
Re: How to find last two digits of a number  
Sir, In solving the problem above, i am assuming that one needs to know the cyclicity for the ten's digit also? So basiclly 3^1 = 01 last two digits 3^2 =09 3^3=27 3^4=81 3^5=43 3^9=83 so after every 4th power the tens digit increases by 4. This would give a cyclicity of 20 for the last two digits? Can this question be attempted if one doesn't follow the cyclicity approach? Also do all numbers have cyclicity of 20 when it comes down to finding the last two digit? 
Re: How to find last two digits of a number  
Hi Abheek Finding last two digits is same as finding remainder of the number with 100. Now if number is coprime with 100 i.e. do not contain 2 or 5, then N^{40} gives remainder 1 with 100 as phi(100) = 1. {I hope you know Euler Theorem of remainders} Also it is found that last two digits of number repeats after every 20 powers. Kamal Lohia 
Re: How to find last two digits of a number  
what will be the last 2 digits of 38 to the power 221? 
Re: How to find last two digits of a number  
Hi, The above question can be solved as follows: (38)^221 = 19^221 x 2^221 Lets look at 19^221= 19^220 x 19 = (19^2)^110 x 19 = Now take the difference of 19 from 50 i.e 31^2 ends in 61 thus, 61^110 = 01 (last two digits) x 19= 19 2^221= As the ten's place of the power is even 24^even =76 x 2^1 ( unit's digit of the power 221). Thus 76x2=152 Now 152x 19 give las two digits as 88. Hope this helps 
Re: How to find last two digits of a number  
Dear Sir, Would it be possible to put up an article on logarithms and binomial expansion? These topics are not as vibrant as you would want them to be on the CAT.but, i feel knowing them well enough would help you skid past those dreadful geometry questions that often circumvent's the mind into the abyss. Ha. Waiting for a reply 
Re: How to find last two digits of a number  
I'm getting answer as 36... Tried twice. 6^77 = (2 * 3)^77 => (2^10)^7 * 2^7 * 3^77 => 24 * 28 * (81)^ 19 * 3 = 72 * 21 * 3 == 72 * 63 = 36 So can you please explain how did you get 56 ? 
Re: How to find last two digits of a number  
tremendous good work 
Re: How to find last two digits of a number  
Thanks!! It was awesome... Whydid I miss your site.. 