IIT JEE , AIEEE Forum - Mathematics , Algebra - Pls ans What are the last two digits of no "9^200"??Plz explain in detail...

towards.destination

Pls ans What are the last two digits of no "9^200"??Plz explain in detail...

aditi_g

write it as (10-1)^200
whn u expand ull get
200 C 0 - 200 C 1 10 + 200C2 100 - 200.............200C200 10^200
now c the first term is 1 second is -2000 third wud be even greater ie.1990000.. and so on.....
now apart frm the first term no other number will have any digit in the ones and tens place...in all other terms it wud be 0 and so wud not contribute...
so the last 2 digits wud be 01..
i hope its clear
lemme knw if u still have any doubt

shubham.123

gud job aditi
i too was thinking the same !

pantpranav

There's another method : The method of observation.

9¹ = 09

9² = 81

9³ = 729

9⁴ = 6561

9⁵ = 59049

9⁶ = 531441

Hence, it is observed that every even power ends in 1.

The digit in ten's place is 8, then 6, then 4 & so on.

So, for 9²⁰⁰ , it should be 0.

Hence the answer is 01.

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