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How to add this ? 1 + 11 + 111 + 9 terms

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How to add this ? 1 + 11 + 111 + 9 terms [#permalink]

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New post 17 Nov 2008, 03:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How to add this ?

1 + 11 + 111 + … 9 terms

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 04:00
I will do it as follows:
1 + 11 + 111 + .....up to 9 terms
= 1 + (10 + 1) + (10^2 + 1) + .......
= 9*1 + (10 + 10^2 + ......+ 10^8)
= 9 + 10(10^8-1)/9

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 04:51
10^2 + 1 =101

but we have 111.

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 06:25
ghentu wrote:
How to add this ?

1 + 11 + 111 + … 9 terms


I am not sure in which format you expect the answer to be, but as long as its 1,11...
answer should be 123456789

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 06:29
echo alpha_plus_gamma.
if the answer choices is the absolute sum, it is pretty easy to visualise and get the answer as
123456789.

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 08:15
Yes .....you got the correct answer .

But how did you solve it ?

please explain

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 08:22
you have 9 terms, so basically the unit digit willl have a sum of 9, because there 9 1's

ten digit will be 8 because there are 8 1's and so on..

its 9 digit sum where the left most number will be 1..

123456789

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 08:58
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ok..for a formulaic approach..here u go...(say the answer is form of \(10^k- xx\))

\(1+11+111...n\)
\(=1/9[9+99+999...]\)
\(=1/9[(10-1)+(10^2-1)+(10^3-1)...]\)
\(=1/9[10+10^2+10^3....-n]\)
\(=1/9[10(10^n-1)/(10-1) - n]\)

here n=9...
\(=1/9[10/9(10^9-1) -9]\)

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 09:48
ghentu wrote:
10^2 + 1 =101

but we have 111.


I think I was feeling sleepy then :oops:

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 10:19
prasun84 wrote:
ok..for a formulaic approach..here u go...(say the answer is form of \(10^k- xx\))

\(1+11+111...n\)
\(=1/9[9+99+999...]\)
\(=1/9[(10-1)+(10^2-1)+(10^3-1)...]\)
\(=1/9[10+10^2+10^3....-n]\)
\(=1/9[10(10^n-1)/(10-1) - n]\)

here n=9...
\(=1/9[10/9(10^9-1) -9]\)



i wish I had known this formula earlier ..

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 15:13
fresinha12 wrote:
you have 9 terms, so basically the unit digit willl have a sum of 9, because there 9 1's

ten digit will be 8 because there are 8 1's and so on..

its 9 digit sum where the left most number will be 1..

123456789



I take the above approach to solve this question as it cannot be better than this one..

I agree that the formula approach is universal for ever case but it is not very helpful in this case.
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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 18:16
Yea, this is probably not meant as a difficult question and it's not realistic to solve that formula during the test within such a short timeframe and without a calculator.


Ghentu, I think what fresinha12 is referring to with visualizing the answer is:

1
11
111
1111
11111
111111
1111111
11111111
+ 111111111
-------------
123456789

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 18:31
nice explanation.

that helped a lot.

thanks

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 17 Nov 2008, 22:58
if u notice theres no formula involved except sum of a GP series...trick is to break it down to a 9 series...the approach wud hold good if it was say 7+77+777...or if the answer is of the form a*(10^k+x)...
fresinha12 wrote:
prasun84 wrote:
ok..for a formulaic approach..here u go...(say the answer is form of \(10^k- xx\))

\(1+11+111...n\)
\(=1/9[9+99+999...]\)
\(=1/9[(10-1)+(10^2-1)+(10^3-1)...]\)
\(=1/9[10+10^2+10^3....-n]\)
\(=1/9[10(10^n-1)/(10-1) - n]\)

here n=9...
\(=1/9[10/9(10^9-1) -9]\)



i wish I had known this formula earlier ..

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 19 Nov 2008, 03:37
prasun84 wrote:
ok..for a formulaic approach..here u go...(say the answer is form of \(10^k- xx\))

\(1+11+111...n\)
\(=1/9[9+99+999...]\)
\(=1/9[(10-1)+(10^2-1)+(10^3-1)...]\)
\(=1/9[10+10^2+10^3....-n]\)
\(=1/9[10(10^n-1)/(10-1) - n]\)

here n=9...
\(=1/9[10/9(10^9-1) -9]\)

prasun, great effort to explain! +1
One question -
How did you reduce expression in the 4th line to the expression in the 5the line?

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Re: 1 + 11 + 111 + … 9 terms = ? [#permalink]

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New post 19 Nov 2008, 03:59
Sum ofa GP Series...
formula is \(a(r^n-1)/(r-1)\)
where a is 1st term, r is the common ratio..

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Re: 1 + 11 + 111 + … 9 terms = ?   [#permalink] 19 Nov 2008, 03:59
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