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Remainder

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vdsoccer
 Post subject: Remainder
 Post Posted: Tue Jan 03, 2012 5:47 am 
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Prospective Students


Posts: 1
What will be the tens digit of :

99^2 + 999^2 + 9999^2........................999999999999^2
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stud.jatt
 Post subject: Re: Remainder
 Post Posted: Wed Jan 04, 2012 8:14 pm 
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Students


Posts: 21
vdsoccer wrote:
What will be the tens digit of :

99^2 + 999^2 + 9999^2........................999999999999^2



split the terms

99^2 + 999^2 +..........

as

(100 - 1)^2 + (1000 - 1)^2 + ........

now (100 - 1)^2 = 10000 - 200 + 1
(1000 - 1)^2 = 1000000 - 2000 + 1

and so on. You don't have to calculate them all, just keep in mind that each one of them has a 0 at ten's digit and 1 at the unit's digit. Thus you get 11 such terms. add them up and the last two digits of the sum will be 11 thus the answer to your question is 1

You can use the same approach for other such series as

98^2 + 998^2 + .......

or

101^2 + 1001^2 + .......
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jnelson0612
 Post subject: Re: Remainder
 Post Posted: Wed Jan 11, 2012 11:07 pm 
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ManhattanGMAT Staff


Posts: 1857
Wow, great answer!

_________________
Jamie Nelson
ManhattanGMAT Instructor
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