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

[vidyarth]

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.

 

[Greg]

Hi vidyarth and welcome to MHB! :D

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

Why?

 

[vidyarth]

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.

 

[Greg]

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

 

[vidyarth]

The answer found [here][1]is $5989\overbrace{\ldots}^{\text {217 9s}}989888$ which is much shorter than your example.
[1]:number theory - Property of 2016 - Mathematics Stack Exchange

 

[Greg]

The answer found ... is $5989\overbrace{\ldots}^{\text {217 9s}}989888$

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

 

[vidyarth]

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

yes. But can you explain how you get it, thoroughly?