IIT JEE , AIEEE Forum - Mathematics , Algebra

sandeepramesh

We strike the first digit of the number $7^{1996}$ and then add it to the remaining number. This is repeated till a number with 10 digits remains. PT this number has 2 equal digits :)

sandeepramesh

no one for this??? I thought this was easy!

vineetnegi

first means unit digit right

sandeepramesh

No not the unit digit Huh?

I mean for 349435345, 3 is the first digit and so on :)

konichiwa2x

If all the digits were different their sum would be equal to

. But we have

and the sum obviouslly remains invariant.

sandeepramesh

nice :) that was the soln i was looking for :D gr8 work koni

vineetnegi

you mean to say

7^(1996)+((7^1996)with first digit striked off) ) and so on

guys I am not sure my ans

but i thought toput it to you

it is 7^(1996)

right

observe the patterm

unit place

7^0= unit place 0

7^1 = 7

patern

1,7,9,3 ,1 so on

so

unit digit in 7^1996

(1996+1)mod4

=1

first in pattern =1

same way for tens digit

pattern is

0,0,4,4,0,0

it can be found that tens place in 7^1996

will be 0

no. of digit in

7^1996 are

1687

there fore to get ten digit left we need to repeat process

1687-10=1677

which means

that

last ten digits get added again and again 1677 times

last digit was ................01*1677

which gives............................77

again i not sure that this is correct or not

sandeepramesh

no i mean to say that 7^{1996} with 1st digit striked off + the digit striked off :)

vineetnegi

oops!

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