Question

Find the sum of the digits of the number 6+66+666+6666 + ... +666...66, where the last number contains 100 digits.

Answer 1

Answer:

Sum of the digits of the number

     $= ((\frac{60((10)^{100}-1) }{81}) - \frac{600}{9} )$

Step-by-step explanation:

Step(i):-

Given series

6+66+666+6666 + ... +666...66 up to 100 digits

Taking common '6'

6 ( 1 + 11 +111+ 1111+1111+.................11111....11 100 digits)

Multiply '9' and divisible by'9'

$\frac{6}{9} X9( 1+11 +111 +1111 +1111+ ..........11111..up to .100 digits )$

Multiply inside '9'

$\frac{6}{9}( 9+99 +999 +9999 +.....+ ..........9999..up to .100 digits )$

$\frac{6}{9}( (10-1)+(100-1)+(1000-1) +(10,000-1)+.....+ ............up to .100 digits )$

$\frac{6}{9}( (10)+(100)+(1000) +(10,000)+.....+ ............up to .100 digits ) - ( 1+1+1+1+........up to 100 digits)$

Step(ii):-

we know that sum of geometric series

$S_{n} = \frac{a(r^{n}-1) }{r-1}$

we know that

a + a + a+........n terms = n a

$\frac{6}{9}( (10)+(10)^{2} +(10)^{3} +(10)^4+.....+ ............up to .100 digits - (100(1))$  ....(i)

The sum of the 100 digits in geometric series

$S_{100} = \frac{10((10)^{100}-1) }{10-1}$

Now the equation (i)

The sum of the digits of the number  

      $= \frac{6}{9} ((\frac{10((10)^{100}-1) }{10-1}) - 100)$

Final answer :-

Sum of the digits of the number

     $= ((\frac{60((10)^{100}-1) }{81}) - \frac{600}{9} )$