What is the smallest three digit number which when divided by 6 leaves a remainder of 5?

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First find the smallest 3-digit number that is evenly divisible by both 5 and 6. Then add 1. Since 5 is prime and not a factor of 6, the smallest positive integer evenly divisible by both 5 and 6 must be 5 times 6 or 30. So the question becomes what is the smallest 3-digit number evenly divisible by 30? 1, 2, and 3 times 30 each produce a 2-digit number. Hence the smallest integer multiplier of 30 that produces a 3-digit product is 4, and the smallest 3-digit integer divisible by 30 is 120. Hence, 120 is the smallest 3-digit integer evenly divisible by both 5 and 6. And finally, 121 is the smallest 3-digit integer when divided by either 5 or 6 leaves a remainder of 1 in each case. John

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 $983$.