Find number with given property

Question

How can we find the smallest $3$-digit number $n$ such that $2 \mid n$, $3 \mid n+1$, $4 \mid n+2$, $5 \mid n+3$ and $6 \mid n+4$ ?

Could you give me a hint?

Answer 1

As @JoelPereira showed, the last digit must be $2$. Moreover, the three digits must add up to a number that is congruent to $2~(\mod 3)$. So we start at the beginning of three digit numbers, i.e., of the form $1x2$ where $x$ is an unknown digit. We search for the first digit $x$ for which $1 + x + 2 \equiv 2~(\mod 3)$ and find $x=2$ works for all the cases.

Thus the answer is $122$.

Answer 2

$\bmod 6\!:\ n\equiv -4\equiv 2$
$\bmod 5\!:\ n\equiv -3\equiv 2$
$\bmod 4\!:\ n\equiv -2\equiv 2$
$\bmod 3\!:\ n\equiv -1\equiv 2$
$\bmod 2\!:\ n\equiv\ \ \ 0\equiv 2$

So the congruences are equivalent to $\,2,3,4,5,6\mid n-2\iff {\rm lcm}(2,3,4,5,6)\mid n-2$

But $\,{\rm lcm}(2,3,4,5,6) = {\rm lcm}(4,5,6) = {\rm lcm}(12,5) = 60,\,$ so $\,n= 2+60k$