4
$\begingroup$

What is the least multiple of $2016$ whose sum of digits is $2016$?

I am totally broke on this. What I only know is that the least number of digits should be greater than $224$ which is clearly not a practical hint. The prime decomposition may give some hint but am unsure. Any ideas. Thanks beforehand.

  • 1
    This is problem J393 [here](https://www.google.at/url?sa=t&rct=j&q=&esrc=s&source=web&cd=17&ved=0ahUKEwjst8a5naPRAhUHUBQKHayxCQE4ChAWCEgwBg&url=https%3A%2F%2Fwww.awesomemath.org%2Fwp-pdf-files%2Fmath-reflections%2Fmr-2016-06%2Fmr_6_2016_problems_1.pdf&usg=AFQjCNGAOQS5TLqvR4uISAW6fxHOV2o8Gg&sig2=KivS_Fun-NhtYAJWkzu2rg). Adrian Andreescu has a book with solutions, I think. – 2017-01-02
  • 0
    @DietrichBurde Definitely. I have taken the problem from there. All the $24$ proposed problems are stupendous. – 2017-01-02

2 Answers 2

4

Looking at the last three digits, you are not going to get more than $24$ digit-sum from them, as the largest digit-sum avilable from multiples of $8$ below $1000$ is at $888$. So you will need $225$ digits in total, which is likely to be enough.

I'll start by finding out the value of $10^{225} \bmod 2016$

$2016=2^5\cdot3^2\cdot7 = 32\cdot 63$

$\lambda(63)$ $= \text{lcm}(\lambda(3^2),\lambda(7)) = \text{lcm}(6,6) = 6 \implies 10^6 \equiv 1 \bmod 63$

$10^{225} \equiv 10^{222}\cdot 10^3 \equiv 10^3\equiv 55 \bmod 63$
$10^{225} \equiv 0 \bmod 32$ trivially

Now Chinese Remainder Theorem says we can find a value that is $ 55 \bmod 63 $ and $0 \bmod 32$ - which translates to looking for a multiple of $32$ that is $\equiv 55 \bmod 63$.

$2\cdot 32 \equiv 1 \bmod 63 \implies (1+2(55-32))\cdot 32 \equiv 47\cdot 32$ is the multiple required, and thus $47\cdot 32 \equiv 1504 \equiv 10^{255} \bmod 2016$

The greatest multiple of $2016$ less than $10^{225}$ is thus $10^{225}-1504 = \overbrace{9\ldots9}^{\text{221 9s}}8496$, which actually does have the digit-sum required. So we already have an upper limit on the minimum value.

Since we have this cycle of $6$ values for $10^k\bmod 2016$ and $10^6 \equiv 1 \bmod 63$ $\implies$ $999999 \equiv 0 \bmod 63$ we can insert "$999999$" into a valid shorter value at any point clear of the last five digits (the factor of $2^5$ being determined by those last five digits). Specifically we can multiply the higher digits by $10^6$ (which doesn't change their value $\bmod 63$) and then add $999999\cdot 10^k$ which of course idivisible by $63$.

Thus when we find the minimum multiple in $9$ digits with a digit-sum of $72$ (which is $598989888$), we can extend that to $225$ digits using insertions of $999999$, keeping $5989$ as the high-value digits.

This gives us $598\overbrace{9\ldots9}^{\text{217 9s}}89888$ as the minimum multiple of $2016$ with a digit sum of $2016$.

  • 0
    This is a great answer but you should probably expand on what that $\lambda $ is doing and why you "need a value that is $55 \pmod{63}$ and $0 \pmod{32}$". – 2017-01-02
  • 0
    I agree that my answer merely gives a way of getting to an easily accessible upper bound. What is required is a practical way of testing lower numbers. Since $2016 = 2^5 \cdot 3^2 \cdot 7$ the final five digits come into question from the factor $2^5$ - it is not immediate to look at the final three, unless you can explain how. It is nice that $999999$ is divisible by $63$ and that divisibility by $32$ depends only on the final five digits. I like what you have done a lot. – 2017-01-02
  • 0
    @gowrath I already linked the first use of the [Carmichael function](https://en.wikipedia.org/wiki/Carmichael_function); essentially finding the minimum (and required) order of exponentiation for a particular modulus, and mentioned CRT again with a link. – 2017-01-02
  • 0
    @MarkBennet thanks, I doubt many people will notice my answer but it was enjoyable putting it together. – 2017-01-02
  • 0
    I didnt understand why you used the carmichael theorem. Also, how did you find the greatest multiple below $10^{225}$ and the process past that is quite unclear. Could you please elaborate? – 2017-01-04
  • 0
    @vidyarthi hope my update answers some of your questions. – 2017-01-04
  • 0
    I don't think I have enough grounding in number theory to understand everything here, but I upvoted it regardless; it appears very well explained. :) I have two follow-up questions: (1) How long did it take you to arrive at this answer (from the time you originally read the question to the time you completed the calculation, even if it wasn't written up yet)? And (2) what book would you recommend to explore number theory for personal enjoyment and deeper mastery than just the very basics? :) – 2017-01-04
  • 0
    Your explanation has clarified the process a little. But, still there are some difficulties. Specifically, in your last but one para, you said "since this cycle of $6$ values for $10^k \bmod 2016$". What is that cycle? And how did you find the minimum multiple in 9 digits with sum 72. – 2017-01-04
  • 1
    @Wildcard it's sometimes hard to judge how long things take, but the first half, to finding the "lucky" $9\ldots98496$, was maybe 10-15 minutes. Then I did a bit of spreadsheet work to find the minimum in $9$ digits so that might have been another 20+ minutes. – 2017-01-04
  • 0
    @vidyarthi successive powers of $10 \bmod 2016$ are $10,100,1000,1936, (1216, 64, 640, 352, 1504, 928)$, with the last six repeating, but I already knew that the cycle existed (without knowing the specific numbers) from the Carmichael function value of $63$ (and knowing that divisibility by $32$ only rests on the last $5$ digits). The minimum in $9$ digits could probably be done elegantly but for me it was easier to just use a couple of spreadsheet calculations. – 2017-01-04
  • 1
    @Wildcard ... and, sorry, I couldn't confidently recommend a book... I just rely on my long-ago mathematics degree and ad-hoc supplemental reading. – 2017-01-04
2

HINT: The sum of the digits of $2016$ is $9$ - can you see a way of exploiting this fact to form a multiple of $2016$ which has some digits equal to $9$? If you can find a multiple with lots of digits equal to $9$ then you may be close to an answer. If you think about this the right way, you should find a way of getting a string of digits equal to $9$.

  • 0
    does multiplying by string of 4s and 5s do? – 2017-01-02
  • 0
    @vidyarthi More complicated than what I was thinking - note that $2+0+1+6=9$ and think of arranging multiples of $2016$ in columns to line up the digits. – 2017-01-02
  • 0
    so the answer is the $111\ldots(224 times)1$ multiple of $2016$, isnt it? – 2017-01-02
  • 0
    @vidyarthi Not exactly. A multiple of $2016$ has to end in one of $0, 6, 2, 4$, or $8$. – 2017-01-02
  • 0
    @gowrath So, my example ends in a $6$ – 2017-01-03
  • 0
    @vidyarthi I may have misunderstood the grammar in your comment. I realize I think you are saying the $111\ldots(224 times)1$-th multiple of $2016$. – 2017-01-03
  • 0
    @gowrath yes, that is what I was saying – 2017-01-03