Last 3 digits

can we convert into 17²⁸.............and do sumthing about it???

how?????????

this is not that tuff or long.....( it is long)
especially since the power is 256........

(300)power......will always end with zeroes.....
so
128C0-128C1(11)..............-128C127(11)¹²⁷(300)+(11)¹²⁸

i think easier way will be taking it as ....(300-11)to d power128....
128C0(300)¹²⁸-128C1(300)¹²⁷(11)........+128c128(11)¹²⁸

I hope u got it?

Sorry if i have wasted ur time if u think it a nit-so-worthy at this point of time sum

sorry for that...

dude.....sorry to point this out............but a small mistake in ur expansion...........ur second last term must be negative and third last positive..............

17 to the powr 256 = (17²) to the power 128
= (290 - 1)to the power 128

= .... .... ... ... ..- 128C125(290³) + 128C126(290²) - 128C127(290) + 1(290) + 1
= 1000k - 128C126(290²) - 128C127(290) + 1(290) + 1

Now calcul;ate bit to get 681 as the answer

i think we will get if we use modulo

ya as sky said......findin last digit is enuf i guess...(at this stage).... but i do want a solution for this......

Well sky, I can find the last 5 digitd but it wud need a bit big multiplications.....

well, were u avle to solve it or not?

U cant follow that method for finding the last 3 digits... that can be done only for the last digit

dun bash me this way guys.... am sry to say it in dat way if it did hurt u all...

but if i ask u abt the 5th last digit... how will u find it ...

well nishant bhiaya told its a 'wat to say' ... not worth-doing question r8 now...

sry once agn...

guys ............for d first few multiples i got d second digits as......(i used a calculator though)..1 8 1 2 5 6 7 4.........no specific trend.....

no question is stupid sky................ and also u dint get d answer for it ok??

Really??? Well It came in RESONANCE AITS last Sunday..... But it isnt a bad question ??? Y do u feel that?/