Notice the pattern:
10² = 100, so 10² - 9 = 91 with a digital sum of 9 + 1 = 10
10³ = 1000, so 10³ - 9 = 991 with a digital sum of 2•9 + 1 = 19
10⁴ = 10000, so 10⁴ - 9 = 9991 with a digital sum of 3•9 + 1 = 28
Then for an arbitrary power of 10, we have
[tex]10^n = 1\underbrace{00\ldots000}_{n\,\rm 0s} \implies 10^n - 1 = \underbrace{999\ldots99}_{n-1\,\rm9s}1 \\\\ \implies \text{digital sum} = 9(n-1)+1 = 9n - 8[/tex]
When [tex]n=51[/tex], the digital sum is 9•51 - 8 = 451.
