Question

Find the sum of place value of last 3 places in the sum of first '10' terms of : 4 + 44 + 444 + ...

Answer 1

$S=4+4+4+...+\mathrm{444....4}\left(10times\right)S=4(1+11+111+....+\mathrm{11..1}(10times)S=\frac{4}{9}(9+99+999+....99..9(10times\left)\right)=\frac{4}{9}(10+{10}^2+{10}^3+....+-10)=\frac{4}{9}({10}^2+{10}^3+....{10}^{10})=\frac{400}{9}(1+{10}^2+.....+{10}^8)=\frac{400({10}^9-1)}{9\times 9}=400\frac{111111111}{9}=400(12345679)$

Sum of place value of last 3 places $=600+0+0$

$=600$