User:Shikha1128

                                          ==  EASY METHOD TO SOLVE LAST DIGIT PROBLEMS ==

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...

2003 x 2004 x 4235161006 x 432657178001 x 42315098002 x 423087004

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.

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 . . . . . . . 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.