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Question Number 199729 by Mingma last updated on 08/Nov/23

Answered by mathfreak01 last updated on 08/Nov/23

$$N=\underset{n=1}{\sum ^{450}}({10}^n-1)$$$$N=\underset{n=1}{\sum ^{450}}{10}^n-450$$$$N=\frac{10({10}^{450}-1)}{9}-450$$$$or$$$$N=(10-1)+(100-1)+...+(1\underset{450digits}{\underbrace{00...00}}-1)$$$$N=\underset{450digits}{\underbrace{11...11}0}-450$$$$N=\underset{447digits}{\underbrace{11..11}1110}-450$$$$N=\underset{447digits}{\underbrace{11..11}0660}$$$$$$

Commented by Rupesh123 last updated on 08/Nov/23

Perfect, sir!

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