IIT JEE , AIEEE Forum - Mathematics , Algebra

SAMADON

the question goes this way

if N=9+99+999+9999+...............+999999....(2007TIMES) .the no. times the digirt 1 is repeated is...........

casio_ss

the given expression can be written as

9( 1+11+111+...(2007 times))

so, the no. of times 1 is repeated will be

n(n+1)/2 = 2007(2008)/2

= 2007* 1004= ans.

chandrasekhar

when a question is put there must be clarity in the question,according to the question which is asked there is no one, if the question is modified it can be done in many ways for example nine can be written as 10 - 1 similarly 99 can be written as 100 - 1 if the question is taken like this digit one is repeated for 2700 times

shreshthmohan

the ans is 11111111111(2003 times).....1111111111109103
so there are 2004 1`s
plz rate

him26.89

the series can be as follows :

9(1 +11+....)=

9(1+1+1+.....................2007times) + 9(10 +100+1000.............2006terms)

=(9 * 2007) +9( 10(10^2006 - 1)/9) [ans]

shine

firstly i wud say
sir is right nd there r many ways in which it can b expressed (u shud mention tht
)if the given expression can be written as

9( 1+11+111+...(2007 times))
then in each tem the no of times 1 is repeated inc by 1
ie like1,2,3,,4................2007
thus the otal no of times r
n(n+1)/2 = 2007(2008)/2
= 2007* 1004= ans.

CooldudeRama17

all u guyz are messing a simple question
look
(10-1)+(10^2-1)+(10^3-1)+(10^4-1)+
the series sum is so simple
10+10^2+.................-n

hmmmmmmmmmmmmm....
hehheh!!

avanti_s.ka

write 99=100-1

999=1000-1

9999=10000-1 n so on

so u have... 100 + 1000 + 10000 +................ -( 1 + 1 + 1......) 2007 times

dats a GP - 2007 .

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