MHB How can the sum of digits of a multiple of 2016 equal 2016?

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The discussion centers on finding the least multiple of 2016 whose digits sum to 2016. It is suggested that this number should be 225 digits long and end in 8, as the minimum number of digits needed to achieve a sum of 2016 is 224, but a configuration of only 9s would not be divisible by 2016. A proposed solution includes a specific number format, which consists of 223 nines followed by 776, but a more concise example is provided as $598\overbrace{9\ldots9}^{\text{217 9s}}89888$. Participants seek clarification on how to derive this solution. The conversation emphasizes the complexity of digit sums in relation to divisibility by 2016.
vidyarth
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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.
 
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Hi vidyarth and welcome to MHB! :D

vidyarth said:
I think the answer must be a 225 digit long number ending in 8 ...

Why?
 
greg1313 said:
Hi vidyarth and welcome to MHB! :D
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.
 
I get 223 followed by 221 9's followed by 776.
 
vidyarth said:
The answer found ... is $5989\overbrace{\ldots}^{\text {217 9s}}989888$

That should be $598\overbrace{9\ldots9}^{\text{217 9s}}89888$
 
greg1313 said:
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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