IIT JEE , AIEEE Forum - Mathematics , Algebra

uday_zingtudor

Hi guys,

Find the last two digits in 2⁹⁹⁹ .Also in 3^999

Thanq

Ronaldooo

For 2^999,is it 68??

sboosy

uday_zingtudor

I know how to find the units place.

Tell me a general procedure for finding tens place

Thanq

hsbhatt

Finding the last two digits is the same as finding the remainder when the number is divided by 100.

You can write $3^{999} = (3^2)^{499} * 3 = (10-1)^{499} * 3$

If you expand $(10-1)^{499}$ using binomial theorem, all the terms will be

divisible by 100 except the terms inom{499}{1} 10 = 4990

and -1

i.e. $3^{998}$ is of the form 100k+4989 = 100k+4900+89 = 100k'+89.

So $3^{999}$ = $3^{998}*3$ is of the form (100k'+89)*3 = 100k"+47.

So 4,7 are in the tens place and units place respectively

rakesh61

Hey i found a short cut methot after going thru these ans see u can find the unit place by using the binomial expansion

by dividing with 10 suppose u get x

as hsbhatt sir has done then to find tens place just subtract from it the base nos 2 and 3 in 2^999 and 3^999 respectively ie x - 2 and x - 3 respectively what do u guys think

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