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Question-199729




Question Number 199729 by Mingma last updated on 08/Nov/23
Answered by mathfreak01 last updated on 08/Nov/23
N = Σ_(n = 1) ^(450) (10^n  − 1)  N = Σ_(n=1) ^(450) 10^n  − 450  N = ((10(10^(450)  − 1))/9) − 450  or  N = (10 − 1) + (100 − 1) + ... + (100...00_(450 digits)  − 1)  N = 11...11_(450 digits) 0 − 450  N = 11..11_(447 digits) 1110 − 450  N = 11..11_(447 digits) 0660
$${N}\:=\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{450}} {\sum}}\left(\mathrm{10}^{{n}} \:−\:\mathrm{1}\right) \\ $$$${N}\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{450}} {\sum}}\mathrm{10}^{{n}} \:−\:\mathrm{450} \\ $$$${N}\:=\:\frac{\mathrm{10}\left(\mathrm{10}^{\mathrm{450}} \:−\:\mathrm{1}\right)}{\mathrm{9}}\:−\:\mathrm{450} \\ $$$${or} \\ $$$${N}\:=\:\left(\mathrm{10}\:−\:\mathrm{1}\right)\:+\:\left(\mathrm{100}\:−\:\mathrm{1}\right)\:+\:…\:+\:\left(\mathrm{1}\underset{\mathrm{450}\:{digits}} {\underbrace{\mathrm{00}…\mathrm{00}}}\:−\:\mathrm{1}\right) \\ $$$${N}\:=\:\underset{\mathrm{450}\:{digits}} {\underbrace{\mathrm{11}…\mathrm{11}}0}\:−\:\mathrm{450} \\ $$$${N}\:=\:\underset{\mathrm{447}\:{digits}} {\underbrace{\mathrm{11}..\mathrm{11}}1110}\:−\:\mathrm{450} \\ $$$${N}\:=\:\underset{\mathrm{447}\:{digits}} {\underbrace{\mathrm{11}..\mathrm{11}}0660} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 08/Nov/23
Perfect, sir!

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