Find the largest 3 digit number which leaves 1 as remainder when divided by 3, 4, 5, 6 and 7

We know one solution $x = 23.\,$ If $x'$ is another solution then $\,x'= 5+6j,\ x= 5+6k\,$ so $\, x'-x = 6(j-k)\,$ is a multiple of $6$. Similarly $\,x'-x\,$ is a multiple of $5$. Thus $\,x'-x\,$ is a multiple of their lcm $= 30.\,$ Conversely if $\,x'-x=30n\,$ then $x',x$ have equal remainders $x$ mod $5$ and $6$.

So the solutions are precisely integers of form $\,23+30n$. The largest multiple of $30$ below $1000$ is clearly $990$ and adding $23$ to that is too big, so we need to subtract $30$ then add $23$, yielding $983$.

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