Find the sum

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Consider the integer

    \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\]

Find the sum of the digits of N

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Let’s express the number in terms of 02bedd46306390e4c9b9293a00c7de6620b71b9b. We can obtain 731c9ca103b784fb2311e097b84f1b0b1ac89dca. By the commutative and associative property, we can group it into 869b85152a42b1318a1510c7b632ac013d94b1cd. We know the former will yield ac7413a564d68b6f152298f6e6a6417a06114811, so we only have to figure out what the last few digits are. There are currently 9691986bc20f256cdd5f2a7c316ddaa8db873e6b 1’s. We know the last four digits are 586f09229b767e5a671847d4894e7cdfb97140d0, and that the others will not be affected if we subtract 9691986bc20f256cdd5f2a7c316ddaa8db873e6b. If we do so, we get that e10ce724df29a726b383c49059ffc910b17d4ef2. This method will remove three dce34f4dfb2406144304ad0d6106c5382ddd1446‘s, and add a e0a0db32027a732ac57d37ef2ae9bb150f65b1088455f3b5cb3b4880b8c9d782a5c1f0334db819eb and bf2c9074b396e3af0dea52d792660eea1c77f10f. Therefore, the sum of the digits is ec55fcb10d3f79b68b44d3f893560ffbe8d9c13c

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