Puzzle #192: Too Many 6’s

This is another beautiful puzzle from A Moscow Math Circle, Week by Week Problem Sets by Sergey Dorichenko. The puzzle will require some manipulation of numbers, but is very solvable and you will be happy that you did it once you get there!

Find the sum

6 + 66 + 666 + 6666 + 66666 + …. + 66…66, if the last string has 100 6’s in it.

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy sixers!

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2 Responses to Puzzle #192: Too Many 6’s

  1. X = 6 + 66 + 666 + 6666 + 66666 + …. + 66…66, if the last string has 100 6’s in it.

    = 6 * (1+11+111+……1…1(100 1s)))
    = 6 * (1*100 + 10*99 + 100*98 + 1000*97 + ……10^100*1)
    = 6 * (1*100 + 10*(100-1) + 100*(100-2) + ….. 10^99(100-99))
    = 6*(100*(1+10+100…..10^99) – (1*10+2*100+…99*10^99))
    = 6*(100*(1+10+100…..10^99) – (1*10+2*100+…99*10^99))
    = 6*(100*S1 – S2)

    where S1 = 1 + 10 + 100… 10^99 = (10^100-1)/9
    and S2 = 1*10+2*100+…. 99*10^99

    10*S2-S2 = 99*10^100 – (10+100 + 1000 ….. 10^99)
    9S2 = 99*10^100 – 10*(1+10…10^98)
    S2 = 11*10^100 – 10 * (10^99-1)/81

    X = 6*[(100/9)*(10^100-1)-11*10^100 + 10 * (10^99-1)/81]
    X = 6*[10^100 (100/9-11+1/81)-100/9-10/81]
    X = 6/81*[10^101 – 910]
    X = 2/27*[10^101 – 910]

  2. Alok Goyal says:

    Pratik, would love to chat on this, I think there is a small error here

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