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

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SUMMARY

The least multiple of 2016 such that the sum of its digits equals 2016 is a 225-digit number ending in 8. The reasoning is based on the fact that the minimum number of digits required to achieve a digit sum of 2016 is 224, derived from dividing 2016 by 9. However, since a number composed solely of 9s is not divisible by 2016, an additional digit is necessary, leading to the conclusion that the number must end in 8. The specific example provided is 598 followed by 217 nines and ending with 888.

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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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