Sum of Digit of A where A is composed of Addends involving 9s Trick
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)
