A number when divided by $902$, $802$ and $702$ leaves remainder of $602$, $502$ and $402$ respectively. What would be the minimum remainder of when the number is divided by $2005$?
My attempt:
For the given number $n$, $n + 300$ is divisible by $902$, $802$ and $702$. Hence $n + 300$ can be written as $\mathrm{lcm}(902,802,702)\cdot K$ where $K$ is the integer and we can keep varying $K$ from $1$ onwards to manually check the minimum remainder. However this is not a great approach. Any simplistic approach to this problem would be appreciated.
Lots of obfuscation.
$n \equiv 502 \pmod {802}$ so $n\equiv 502 \equiv 101 \pmod {401}$.
So $n =101 + 401K$ for some $K$ so $\pmod {2005 = 401 \times 5}$ we have $n \equiv 101 + 401k$ for $k = 0,1,2,3,4$
So $n \equiv 101 + 401k \pmod {2005}$. The least that can be is $101$ if $k=0$ and we'd have $n \equiv 101 \pmod 5$.
Now if we want
$n\equiv 602 \pmod{902}$
$n \equiv 502 \pmod {802}$
$n \equiv 402 \pmod {702}$ and
$n \equiv 101 \pmod 5$ we just need
$n \equiv 0 \pmod 2; n \equiv 602 \pmod 451; n \equiv 502 \pmod{401}; n\equiv 402\pmod{351}; $ and $n\equiv 101 \pmod 5$.
And as $2, 451, 401, 351, 5$ are relatively prime. So by CRT we can have that.
(And thank god we don't have to find that...... well, it wouldn't be that bad[1] but... we don't have to find it; it's enough to know it exists)
And so $n \equiv 101\pmod 5$ and $n \equiv 101 \pmod {401}$ and so $n\equiv 101 \pmod {2005}$ and $101$ is the smallest possible remainder.
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[1] If we actually want to solve:
$n \equiv -300 \pmod {902, 802, 702}$ so $n\equiv -300 \pmod{2;451;401;351}$ so $n \equiv -300 \pmod {126957402}$ and $n \equiv 1\pmod 5$ so $n\equiv -300 + 126957402k \pmod {634,787,010}$ where $k = 0,1,2,3, 4$ but as $n \equiv 1 \pmod 5$ we must have $k=3$ so $n \equiv 380871906 \pmod {634,787,010}$ and indeed $380871906 \pmod {2005} \equiv 101$.
And $380871906 \equiv 602;502;402 \pmod{902;802;702}$.