__Type # 1:__

questions of the form…

2003 x 2004 x 4235161006 x 432657178001 x 42315098002 x 423087004

**concept:**

**Last n digits of any product depends on the product of last n digits. so just multiply last n digits of each term…find the product, take last n digits of the product n multiply it with the next term…continue this for all terms.**

suppose v had to find last digit of the product above…

so multiply last digits.

3 x 4 = 12. dont worry abt 1 in 12. just remember 2 and multiply it with next no. 2 x 6 = 12. so 2, 2 x 1= 2, 2 x 2 = 4, 4 x 4 = 16.

so, the ans is 6.

for last 2 digits:

03 x 04 x 06 x 01 x 02 x 04 = 76

for last 3 digits:

003 x 004 x 006 x 001 x 002 x 004 = 576.

trust me, last 4 digits wont be asked…as then it becomes bulky…questions in cat are tricky…

__Type # 2:__

of the form…last digits of 432^43567. i.e base ^ power

i know most of us know this concept…wud take it concisely…

look for the variation in last digit of higher powers of last digit of the base…i.e. 2 here.

2^1 = >2

2^2 => 4

2^3 => 8

2^4 => 6

2^5 => 2

so we can say that 2,4,8,6 will keep repeating…no. of different digits that the last digits of higher powers can take is known as cyclicity. every digit has a cyclicity.

to find last digit, find last digit of :

**(last digit of base) ^ (power % cyclicity of last digit)**

when, power % cyclicity of last digit = 0,

take,

**(last digit of base) ^ (cyclicity of last digit)**

in the above example…

432^43567.

2^ (43567 % 4) = 2^3 = 8 answer.

with little practice….u can do all such questions mentally.

__Type # 3__

Find last 2 digits of 34291^201.

when the last digit is 1,9,0,5…try finding a pattern in last 2 digits…u’ll get one…then solve the question accordingly…

last 2 digits of 91^1 = 91

last 2 digits of 91^2 = 81

last 2 digits of 91^3 = 71

last 2 digits of 91^4 = 61

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last 2 digits of 91^10 = 01

last 2 digits of 91^11 = 91

see, we again got 91 as last 2 digits…so v can say that the cyclicity of 91 for last 2 digits is 11 -1 = 10

so the answer shud be 91.

lets take one more example…

find last 2 digits of

(49)^(37)^(38 )^(39)…(3700)

see…

last 2 digits of 49^1 => 49

last 2 digits of 49^2 => 01

last 2 digits of 49^3 => 49

last 2 digits of 49^4 => 01.

can v say that for all odd powers, answer wud be 01?

yes!

since the power of 49 is odd…the answer shud be 01.

what if the question was

(49)^(37)^(38 )^(39)…(3700)%100 = ?

answer wud be 01%100 = 1

(49)^(37)^(38 )^(39)…(3700) % 20 = ?

answer wus be 01 % 20 = 1.

bcoz, to c remainder with 20, v need last 2 digits only.

(49)^(37 )^(38 )^(39)…(3700) % 10

=> 49 % 10 = 9

this method might become tedious when the last 2 digits are unfriendly…like 37, 82 etc. but i have neva seen such figures appearing in cat…to solve such figures, v need euler’s or binomial…ill be taking it when v discuss remainders…today’s questions are based on the above concept only..

**Questions for today..**

**1. last digit of 3677^400 – 689^84**

**2. last digit of 11^11 + 12^12….1000^1000.**

**3. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451…do i need to say….its tricky…not lengthy!!**

**4. no. of zeroes at the end in 1^1×2^2×3^3x…250^250.**

**5. no. of zeroes at the end in 1! x 2! x 3! x…25!**

**6. Last non-zero digit of 25!**

**7. Last non-zero digit of 1! x 2! x 3!…15!**

Find **last 2 digits of…**

**8. 81^(371)^(372)^…(400)**

9. **11 ^ (25)^(31)^(41)…(1001)**

10. **7 ^ 2501.**

**11. 3^2537837.**

September 18, 2008 at 11:20 am

Hi Bhavin,

Nice explanation/post. Do you also post the anwers somewhere?

If not can yuo send me teh answers for this excercise.

Cheers,

Ekta

September 23, 2008 at 8:54 am

REGRETS for not putting the solutions…

1. last digit of 3677^400 – 689^84—- 0

2. last digit of 11^11 + 12^12….1000^1000

Ans 3

from 1^1 till 10^10, we’ll get 1,6,3,6,5,6,7,4,9,0. the order will change but the set will be of these 10 digits only…just an observation!!

so, we’re adding the bove sum (1000-10 ) /10 = 99 times…so the last digit shud be last digit of 47 x 99 = 3.

3. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451

Ans – 50

there’s one 25 n one 2…that product will make the last 2 digits 50. now 50 x any odd no. will always leave 50 as last 2 digits…n all the other nos. are odd

4. Last non-zero digit of 25!

Ans – 5

keep multiplying last digits…not a matter of more than 90 secs.

5s… 5 ( 1+ 2+….50) = 6375

25s…25(1+2+..10) = 1375

125s…125(1+2) = 375

total = 8125

5. no. of zeroes at the end in 1! x 2! x 3! x…25!—-56

5-9…5 nos. one 0 each = 5

10-14…5 nos. two 0 each=10

15-19…5 nos. three 0 each=15

20-24…5 nos. four 0 each=20

25…1 no. with 6 0s = 6

total = 56

6. Last non-zero digit of 25!

Ans – 4

7. Last non-zero digit of 1! x 2! x 3!…15!

ans 6

same approach as above

Find last 2 digits of…

8. 81^(371)^(372)^…(400)—81

9. 11 ^ (25)^(31)^(41)…(1001)—-51

10. 7 ^ 2501.—-07

11. 3^2537837.—-63

Eular No. for 100 is 40, so

3^2537837 mod 100 —> 3^(-3) mod 100

=>(((3*33)^(-3))*(33^3))%100=> ((-1)(33^3))%100=>-37 (mod 100)

-37 (mod 100)=>63 (mod 100)

July 2, 2010 at 4:10 pm

hey bhavin

can me give full explanation of q. no 2,4,5,7

November 1, 2008 at 6:20 am

hi……….. can u help me for finding last non zero digit of n! here n can be any 2 or 3 digit number.

December 25, 2008 at 10:50 am

4!

January 4, 2009 at 12:33 pm

4!

= 24 so need not solve through difficult method.

August 29, 2009 at 8:20 pm

hi if a number is of the format a^b where a is not an integer then how will u calculate the last m digits before the decimal point

February 5, 2013 at 8:24 pm

Ignore the decimal point.

September 15, 2009 at 12:06 pm

i want to know the last 10 digits in the series

1^1 + 2^2 + …………………+1000^1000;

THANKS FOR YOUR HELP;

February 5, 2013 at 8:21 pm

Wolframalpha computes 10 last digits of that: …9110846700. Copy the link and check here: http://www.wolframalpha.com/input/?i=sum+n%5En%2C+n%3D1+to+1000

September 23, 2009 at 6:33 pm

Found this very helpful. Thanks!

September 23, 2009 at 6:36 pm

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November 18, 2009 at 12:53 pm

Can any one give the solutions with full explanation.

Thanks in advance……

May 7, 2010 at 5:03 pm

Hi can you help me finding the last digit of the power 2^1001 thanks ………..

February 5, 2013 at 8:23 pm

This is 2. Check the big calculator to verify.

June 28, 2011 at 1:55 pm

pl tell me how to find the last digit of 13 raised to the power 2003

February 5, 2013 at 8:27 pm

This digit is 7, if you consider the sequence of 13^1, 13^2, 13^3, you find that all 13^3, 13^7, 13^x (where x+1 div 4 = x+1/4) have 7 in the last digit.

April 30, 2013 at 5:25 am

I WANT EXPLANATION FOR LAST DIGIT OF 38^2011 IS 2

June 22, 2013 at 2:15 pm

this digit is 2, if you consider the sequence of 8^1=8,8^2=4,8^3=2,8^4=6 so 8^(2011%4)=8^3=2

June 23, 2016 at 5:00 pm

Can u tell me…

What is the .. last digit of. 13^14^15^16^17^18^19^………^n..

August 19, 2016 at 9:08 am

This is nostalgic … the last time i solved the problem was in 2008.