Finding the largest remainder when $2019$ is divided by $1$, $2$, $3$, $\ldots$, $1000$ 
If I have $2019$, and I divide $2019$ successively by $1, 2, 3$, and so on, up to and including $1000$, how would I find out the largest remainder?

I felt like dividing all the numbers each would take too long, and I thought there should be some kind of method to solve this kind of question. I do not know how to work this out, and the strategy behind it.
 A: We have $2019= 3 k$ where $k=673$. When dividing by numbers from $k+1$ to $1000$ the quotients will be the same, $2$, so the remainder will be smaller when the divider increases. Dividing by $k+1$ we get
$$3 k = 2(k+1) + k-2$$
so the largest remainder when dividing at numbers from $k+1$ to $1000$ is $k-2$.
When dividing by $k-1$ we get
$$3k = 3(k-1) + 3$$
so remainder $3< k-2$. When dividing by numbers from $1$ to $k-2$ we get remainders $< k-2$. Therefore, the largest possible remainder is $k-2 = 671$, obtained when dividing by $k+1 = 674$.
A: By the division algorithm, $n=dq_d+r_d$, ie, $r_d=n-dq_d$. To maximize $r_d$, we need to minimize both $d$ and $q_d$.
Given the allowable range of $1\le d\le 1000$, the least we can minimize $q_d$ is $q_d=2$ because for $q_d=1\lt 2$, we would need $d\gt\lfloor n/2\rfloor=1009$ which is $\gt 1000$, the allowed range.
So, we can minimize $q_d$ to $2$ which gives us the range $\lfloor\frac{2019}3\rfloor\lt d\lt\lfloor\frac{2019}2\rfloor$ to choose $d$, ie, $673\lt d\lt 1009$; minimize with $d=674$ and the maximum possible remainder is $r_{674}=671$
A: Intuitively dividing $2n + 1$ by $n+1$ will give remainder $n$.
Dividing by anything less than or equal to $n$ will give a remainder less than $n$.
Dividing by any $m > n+1$ will give a remainder of $(2n+1) - m < (2n+1)-(n+1) = n$.
So $2019 \pmod {1010} \equiv 1009$ and .... oops your answer requires we divide by less than or equal to $1000$.
Okay if we divide $N$ by $\lfloor \frac Nk \rfloor + 1$ we get....
.... well if $\lfloor \frac Nk \rfloor = q$ if and only if $qk \le N < (q+1)k$.
So if we divide by $q+1$ we get $(k-1)(q+1) =kq -q+k-1$ and if $q > k$ (in other words of $k< \sqrt N$) then $(k-1)(q+1)=kq-q+k -1 < qk \le N < (q+1)k$ so
the remainder will be $N-(k-1)(q+1) > qk -(kq-q+k-1)=q-k+1$.
We can maximize that by make $k$ really small and $q$ really big.
If we make $k =2$, and $q = 1010$ as I did at the top of the post that is the biggest remainder.  But $q=1010$ is to big to divide by.  So let $k =2$.  $3\times 673 = 2019$ so let $q = 674$.
THen $2\times 674 < 2019 < 3\times 674$ and the remainder is $2019 - 2\times 674 =671$.
For any $n=673$ or $672$ you get remainders $0$ and $3$ when dividing by $n$.  $2019 = 3\times 673$ and $2019 = 3\times 672 + 3$.  For $n \le 671$ you get a remainder less than $n$ and so less than $671$.
And for any $n = 674 + m$ where $1\le  m \le 1000-673= 327$ we have $2(673 + m) < 2000 < 2019 < 3(673+m) = 2019 +m$ so the remainder is $2019 - 2(674+m) = 671 - 2m < 671$.
So $671$ is the largest remainder.
