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

Click For Summary

Discussion Overview

The discussion revolves around finding the least multiple of 2016 such that the sum of its digits equals 2016. Participants explore the properties of digit sums and the structure of such numbers, considering both theoretical and numerical aspects.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes that the answer must be a 225-digit long number ending in 8, although they are unsure of the exact value or proof.
  • Another participant suggests that the minimum number of digits needed to achieve a digit sum of 2016 is 224, based on the calculation of 2016 divided by 9, but notes that a number consisting solely of 9s would not be divisible by 2016, thus requiring an additional digit.
  • A different participant claims to have found a shorter number, specifically $598\overbrace{9\ldots9}^{\text{217 9s}}89888$, which they argue is much shorter than the previously suggested example.
  • There is a request for a thorough explanation of how the shorter number was derived, indicating a desire for deeper understanding of the reasoning behind the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the length and structure of the number that meets the criteria, with no consensus reached on the exact form or proof of the solution.

Contextual Notes

Some assumptions regarding the divisibility of numbers and the properties of digit sums are not fully explored, leaving open questions about the methodology used to derive the proposed numbers.

vidyarth
Messages
16
Reaction score
0
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.
 
Mathematics news on Phys.org
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?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
7
Views
9K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 3 ·
Replies
3
Views
4K