SASMO Question in Image below

Solution

21) Notice that

9 = 10 – 1

99 = 100 – 1

999 = 1000 – 1

.

.

.

999…999 (2017 digits) = 10^2017 – 1

So

A = (10^1 + 10^2 + 10^3 + … + 10^2017) – (1 + 1 + 1 + … + 1 with 2017 one’s)

A = (10^1 + 10^2 + 10^3 + … + 10^2017) – 2017

Method I (Geometric Series Formula)

The sum

(10^1 + 10^2 + 10^3 + … + 10^2017)

is a Geometric Series with first term 10 and common ratio = 10

The formula for the sum of the Geometric Series is as above where

S = Sum of the first n terms

a = a(1) = First Term = 10

r = common ratio = 10

n = number of terms = 2017 terms

(10^1 + 10^2 + 10^3 + … + 10^2017) = [a (r^n – 1)] / [(r – 1)]

Becomes

(10^1 + 10^2 + 10^3 + … + 10^2017)

= [a (r^n – 1)] / [(r – 1)]

= [10 (10^2017 – 1)] / [(10 – 1)]

=   [10 (10^2017 – 1)] / 9

please note that

(10^2017 – 1) = 1000…0000 (2017 zeroes) – 1 = 999…999 (a number consisting of 2017 nine’s)

Thus from the above

(10^1 + 10^2 + 10^3 + … + 10^2017)

=   [10 (10^2017 – 1)] / 9

= 10 x 999…999 (a number consisting of 2017 nine’s) / 9

= 10 x 111…111 (a number consisting of 2017 one’s)

= 111…1110      (a number consisting of 2017 one’s followed by a zero)

Now let’s compute A

Recap

A = (10^1 + 10^2 + 10^3 + … + 10^2017) – 2017

So,

A =  111…1110      (a number consisting of 2017 one’s followed by a zero)   – 2017

Note that the last five digits of the first addend subtracts 2017 —-> we consider

11110 – 02017 = 09093     (the zero is important because the subtraction is happening to a larger number – place value, positioning)

What does it mean?

The other digits of “1” remain intact and followed by the last four digits above (9093)

A= 111…11109093 (a number consisting of 2012 ones followed by these five last digits 09093)

Sum of Digits of A

= ( 2012 ones) + 0 + 9 + 0 + 9 + 3

= 2012 + 21

= 2033 (answer)