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 life-saving 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
b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2
d. Last two digits of numbers which end in 4, 6 and 8

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 31⁷⁸⁶?

Solution: 31⁷⁸⁶ = (30 + 1)⁷⁸⁶ = ⁷⁸⁶C₀  ×— 1⁷⁸⁶ + ⁷⁸⁶C₁  ×— 1⁷⁸⁵ ×— (30) + ⁷⁸⁶C₂ ×— 1⁷⁸⁴ ×— 30² + ..., Note that all the terms after the second term will end in two or more zeroes. The first two terms are ⁷⁸⁶C₀  ×— 1⁷⁸⁶ and ⁷⁸⁶C₁  ×— 1⁷⁸⁵ × (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 31⁷⁸⁶ are 81.

Now, here is the shortcut:

 
Here are some more examples:

Find the last two digits of 41²⁷⁸⁹

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 71⁵⁶⁷⁴⁷

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 51⁴⁵⁶  ×— 61⁵⁶⁷

The last two digits of 51⁴⁵⁶ will be 01 and the last two digits of 61⁵⁶⁷ will be 21. Therefore, the last two digits of 51⁴⁵⁶  ×— 61⁵⁶⁷ 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 19²⁶⁶.

19²⁶⁶ = (19²)¹³³. Now, 19² ends in 61 (19² = 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 33²⁸⁸.

33²⁸⁸ = (33⁴)⁷². Now 33⁴ ends in 21 (33⁴ = 33² ×—33² = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21⁷². By the previous method, the last two digits of 21⁷² = 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 87⁴⁷⁴.

87⁴⁷⁴ = 87⁴⁷² ×—87² = (87⁴)¹¹⁸ ×—87² = (69 × 69)¹¹⁸ × 69 (The last two digits of 87² are 69) = 61¹¹⁸ ×— 69 = 81 ×— 69 = 89

If you understood the method then try your hands on these questions:

Find the last two digits of:

1. 27⁴⁵⁶
2. 79⁸³
3. 583⁵¹²
 

Last two digits of numbers ending in 2, 4, 6 or 8

There is only one even two-digit 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 24² ends in 76 and 2¹⁰ 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, 24³⁴ will end in 76 and 24⁵³ will end in 24.

Let's apply this funda:

Find the last two digits of 2⁵⁴³.

2⁵⁴³ = (2¹⁰)⁵⁴ ×—2³ = (24)⁵⁴ (24 raised to an even power) ×—2³ = 76 × 8 = 08

(NOTE: Here if you need to multiply 76 with 2ⁿ, then you can straightaway write the last two digits of 2ⁿ because when 76 is multiplied with 2ⁿ the last two digits remain the same as the last two digits of 2ⁿ. Therefore, the last two digits of 76 × 2⁷ will be the last two digits of 2⁷ = 28)

Same method we can use for any number which is of the form 2ⁿ. Here is an example:

Find the last two digits of 64²³⁶.

64²³⁶ = (2⁶)²³⁶ = 2¹⁴¹⁶ = (2¹⁰)¹⁴¹ × 2⁶ = 24¹⁴¹ (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 62⁵⁸⁶.

62⁵⁸⁶ = (2 × 31)⁵⁸⁶ = 2⁵⁸⁶ × 3⁵⁸⁶ = (2¹⁰)⁵⁸ ×— 2⁶ × 31⁵⁸⁶ = 76 × 64 × 81 = 84

Find the last two digits of 54³⁸⁰.

54³⁸⁰ = (2 × 3³)³⁸⁰ = 2³⁸⁰ × 3¹¹⁴⁰ = (2¹⁰)³⁸ × (3⁴)²⁸⁵ = 76 × 81²⁸⁵ = 76 × 01 = 76.

Find the last two digits of 56²⁸³.

56²⁸³ = (2³ × 7)²⁸³ = 2⁸⁴⁹ × 7²⁸³ = (2¹⁰)⁸⁴ × 2⁹ × (7⁴)⁷⁰ × 7³ = 76 × 12 × (01)⁷⁰ ×—43 = 16

Find the last two digits of 78³⁷⁹.

78³⁷⁹ = (2 × 39)³⁷⁹ = 2³⁷⁹ × 39³⁷⁹ = (2¹⁰)³⁷ × 2⁹ × (39²)¹⁸⁹ × 39 = 24 × 12 × 81 × 39 = 92

Now try to find the last two digits of
1. 34⁵⁷⁶
2. 28²⁸⁷

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.

If you think this article was useful, help others by sharing it with your friends!

hi if problems last digit is 5 like 25⁴⁵

^{den wat will be last digits}

1. Numbers where the previous digit of 5 is 0 or any even number and
2. Numbers where the previous digit of 5 is any odd number

2^2^2008 will not translate into 4^2008.

Just to clarify, check the value with 2^3^2

Logically, it should translate into 2^9

But, if we go by your approach, it would translate into 8^2....

IMHO,

2^2^2008,  first, lets find the last 2 digit of 2^2008.

last two digit will be 56,

now, we have a number which is 2^xxxx56

it will translate into  (2^10)^odd number)*2^6

=(24)^odd number *2^6=24*64=36

Thanks Ashish for such useful fundas..

Hi ,

Thanks for the nice article!!!!

What are the ans for-:

Find the last two digits of:

1. 27⁴⁵⁶
2. 79⁸³
3. 583⁵¹²
 

Is it

1. 89

2. 69

3.  61

Please reply

Thanks

Could anybody help with problems of the type
2007^2008^2009 please???

soln..

cyclicity fr last 2 places fr 7 is 4..

cycle is

07

49

43

01

nd in d power we hv multiple of 4..so, ans will be 01.

finding last two digits of a number ending wd 5..

(.......E5)^even = 25, where E represents even no.

                       75, otherwise

Hi ,

As always with TG articles, a very good article! Thanks Ashish and TG

Can someone explain the following? How to use this ? Please explain with some examples.

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

Find the last two digits of:

Dear Ashish,

Thank you very much for this valuable post.

Regards.

Hello

These methods are just awsome. time saving, fast and verbal calculation.

Great Job Done.

Hello Sahana

I have got 01 as the answer of this Problem. Is this Right ?

What if we cud have 2003^2004^2005 and so on....

and need to find last 2 digits.....

Hi can anyone please tell me the detailed Solution for this Question.

2003^2004^2005.....

LAST 2 DIGITS

Hi Netra,

Answer is Abstly Correct. But Dont you think This Method is Quite Long..

1st Step is Fine abt 5  power 25 all tht. 2nd we can do (2000+4) power.............25. 2000 wl gv 00 so we js need 2 check 4 ( as u know).

2 power.........50. will gv 24 as last digit as u explained.

den 2000+3) power 24. We js need to consider 24 as we can get Lst 2 digits wid tht. 3 power 24. OR 81 power 6.

so it will gv 81 As the Last 2 digits.

Am i Right?

Pallav

Hello Guys,

Is there any one interested in Material Swaping. Who think that he or she have complete their Material so that we can Swap our Material for Some Time..

Reply.....

Pallav

Hi Pallav,

I think we all have gone too far with it, here's what i did....

For 2003^2004^2005 expression we need to calculate the last two digits, hence the last two digits for it will be the same as the last two digits for 03^04^05 or 3^4^5 for which the last two digits come up as 81.

Regards

Ashish 

Ashish,

I cannnot think of a simpler approach then you have suggested. I don't have the words to say thanks. Your contribution will be remembered by all CAT aspirants whjo read this article

Hi SE,

can u pls elaborate how to solve this question (last two digits of 2003^2004^2005) with the help of euler's theorem. i am getting 43 as the answer by that method..small pointers of the solution i did

phi(100) = 40 then solving further, i get Remainder[(2003^40k +5)/100]

which gives 43

pls elaborate. also i have read whole of the ur article on the remainders.. and religiously solved each n every question by my own and too using just eulers theorem, and i ahve some more doubts will post in that article only. thanks!!

waiting for the reply of the query asked above.

ultimate!!

cheers!!

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...

GR8 wrk.... i admire u 4 such wondrful work.......                                    kudos 2 u....                                                                                          keep it up....

Find the last two digits of:

What if we cud have 2003^2004^2005 and so on....

and need to find last 2 digits.....

My Dear

Last 2 digit for any power of 5 will always 25.

Try it...

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.....

Bang on the target!!!!!!!!!!

Must say Fundooooooooo article TG.......

U rock!!!!!

hey.......please tell me how did you get that....

last digit 56.....

please elaborate it......

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??

1. 27^456 = 61
2. 79^83  = 39
3. 583^512 = 61
4. 34^576 = 16
5. 28^287 = 12

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
a) 4
b) 6
c) 8
d) 0

Ans - (a)

2.) [x] denotes the greatest integer value just below x and {x} its
fractional value.The sum of [x]^2 and {x}^1 is 25.16.Find x
a)5.16
b)-4.84
c)Both (a) and (b)
d)4.84

Ans - (c)

3.) The power of 45 that will exactly divide 123! is
a)28
b)30
c)31
d)59
e)29

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
as the first is lessser than the second.If the product of the two smaller
number is 85 and the product of two larger number is 115 find the middle
number.
a)9
b)8
c)12
d)30
e)10

Ans - (e)

Thanks

Rahul

Hi Rahul

1) (12⁵⁵/3¹¹) + (8⁴⁸/16¹⁸) = (3⁴⁴)(4⁵⁵) + 2⁷² = (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 = 3²*5 and 123! contains 3^41+13+4+1 = 3⁵⁹ or 9²⁹ and 5²⁴⁺⁴ = 5²⁸ so 45²⁸. (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) = 2b² = 85 + 115 = 200 = 2*10². So b = 10 (e)  

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

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.

Hi Bhaskar

79⁸³ = (79)(79)⁸² ≡ (79)(21)⁸² mod100 ≡ (79)(41) mod100 ≡ 39 mod100.

Kamal Lohia

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.

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.

Awesome article as well as Comments. Both helped me a lot.

Cheers

superbbbb

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?

Multiply the last 2 digits of the numbers ie, 89*89 = 7921 so last 2 digits of1089*1089 is 21.

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 

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 (2-digit 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

Hi ,this is really helpful.

thanks Gowtham.

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 co-prime, so 14²⁰ = 1 mod 25.

Also 14¹⁴ = 20k + 16

=> 14^{14^14} =14¹⁶ mod25 = 4⁸ mod25 = 2¹⁶ mod25 = -(2⁶) mod25 = -64 mod25 = 11 mod25 = 36 mod25.

As number is divisible by 4, its last two digits are 36.

For: last two digits of 14^14^14??
Where am i going wrong?

-> 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?

Hi Rhythm Goyal

Everything was fine except second-last step where it should be 24 in place of 76 and the answer will be 36.

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?

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)

1. i think the answer should be 39