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

    Anonymous Christian
    Mar 28, 2019
    Math Olympiad, Math Olympiad Secondary, SASMO
    Comments Off on 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)

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