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  1. MHB Apprentice

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    #1
    What is the least multiple of 2016 such that the sum of its digits is 2016.
    I think the answer must be a 225 digit long number ending in 8 but do not know the exact value nor how to prove it. Any ideas. Thanks beforehand.

  2. Perseverance
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    #2
    Hi vidyarth and welcome to MHB!

    Quote Originally Posted by vidyarth View Post
    I think the answer must be a 225 digit long number ending in 8 ...
    Why?

  3. MHB Apprentice

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    Quote Originally Posted by greg1313 View Post
    Hi vidyarth and welcome to MHB!



    Why?
    This is because the least number of digits required to get a digit sum of $2016$ is $\frac{2016}{9}=224$. But since a string consisting of only $9$s is not divisible by $2016$, therefore it can be made up by using one extra digit. And I think the number should end in $8$ which is the second maximum digit.

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    #4
    I get 223 followed by 221 9's followed by 776.

  5. MHB Apprentice

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    #5 Thread Author
    The answer found [here][1]is $5989\overbrace{\ldots}^{\text {217 9s}}989888$ which is much shorter than your example.
    [1]:

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    #6
    Quote Originally Posted by vidyarth View Post
    The answer found ... is $5989\overbrace{\ldots}^{\text {217 9s}}989888$
    That should be $598\overbrace{9\ldots9}^{\text{217 9s}}89888$

  7. MHB Apprentice

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    #7 Thread Author
    Quote Originally Posted by greg1313 View Post
    That should be $598\overbrace{9\ldots9}^{\text{217 9s}}89888$
    yes. But can you explain how you get it, thoroughly?

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