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How can I find a minimal natural number that begins (in the decimal system) with $2016$ and which can be divided by $2017$?

3 Answers 3

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I computed
$2017-\bmod(20160,2017)=10$,
$2017-\bmod(201600,2017)=100$,
$2017-\bmod(2016000,2017)=1000$,
$2017-\bmod(20160000,2017)=1932$.

Therefore, I get $$ 20161932=2017\cdot9996 $$

3

It must start with the numbers $2016...$, so it must be of the form $$2016 \times 10^{n}+m $$

Where $m<10^{n}$. Now note $$2016 \times 10^{n}+m \equiv -10^{n}+m \equiv 0 \pmod {2017}$$ So, we have that $10^{n}-m$ must be a positive multiple of $2017$.

So $10^{n}-m=2017 \times 4$ is the method to minimize the integer. We get $n=4, m=1932$ The number is $$20161932$$

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    I get 20161932. – 2017-02-26
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    How did you get it? How do you get the exact values of n and m? – 2017-02-26
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    Why not $m=1932$? $20161932=2017\cdot9996$. – 2017-02-26
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    @robjohn can you share an answer? – 2017-02-26
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    This answer is trivially incorrect because you can substract $2017$ to that number and still have a number in the form $2016xyzw$ and divisible by $2017$ – 2017-02-26
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    @Giulio Sorry, I will rectify my answer accordingly. – 2017-02-26
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    @robjohn I edited my answer. – 2017-02-26
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    @HenningMakholm I edited my answer. – 2017-02-26
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    @Peter I edited the answer. – 2017-02-26
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    @idliketodothis Sorry for my previously incorrect answer. – 2017-02-26
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    The first equivalence is modulo 2017, right? – 2017-02-26
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    @idliketodothis Yes. That is true. That's why it's written $\pmod {2017}$. – 2017-02-26
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    It isn't. I am talking about that one: 2016×10n+m≡−10n – 2017-02-26
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    @idliketodothis Ohk. Fixed that. – 2017-02-26
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It is clear that there must be an answer of the form 2016xxxx because the 10,000 consecutive numbers of this form are enough that there are several multiples of $2017$ among them.

We can also see that 2016xxx, 2016xx, or 2016x won't work, because each of these gives a range of fewer than 2017 numbers that end just before a multiple of $2017$.

Finding the first solution of the form 2016xxxx is just a matter of rounding $20{,}160{,}000$ up to the next multiple of $2017$: $$ 2017 \times \left\lceil\frac{20160000}{2017}\right\rceil = 20161932 $$