Question

sum of digit of a number 10^n -1 is 3798 then the value of n is

Answer 1

${10}^{n} - 1$
Will be of the form 9999......... and therefore the sum of the digits will be 9 times the number of 9'ns. Thus, the number of nines is
$\frac{3798}{9} = 422$
Thus
${10}^{n} - 1$
Contains 422 nines. But since when subtracting 1 from 10^n, 1 gets carried over,
$n = 422 + 1 = 423$
Brainliest!!!!