Increasing remainders – 2

$n \bmod 1 = 0$ for all integers.

$n \bmod 2 \in \{0,1\}$ , but it can't be 0 because the condition requires $n \bmod 1 < n \bmod 2$ , so $n \bmod 2 = 1$ .

$n \bmod 3 \in \{0,1,2\}$ , but again, the condition leaves only 2, so $n \bmod 3 = 2$ .

Following the same logic, we can deduce $n \bmod b = b-1 \Leftrightarrow n+1 \bmod b = 0$ up to $b=9$ . So, let's find out the smallest $n+1$ . Since it leaves a remainder of 0 under division by $1,2,3,\ldots,9$ , it has to be the least common multiple of these numbers, which is $2^3 \cdot 3^2 \cdot 5^1 \cdot 7^1 = 2520$ and therefore $n = \boxed{2519}$ .

Let us start with $1$ . We know $n$ mod $1=0$ . Now the remainder when n is divided by $2$ must be $1$ . As $n$ mod $b< n$ mod $(b+1)$ and remainder when $n$ is divided by $2$ is at most $1$ . Thus, the sequence of the remainders when n is divided by $1,2,.....,8,9$ is respectively $0,1,.....8$ . Hence, we can write $n=p*b+b-1$ . Or, $b|n+1$ . So, the minimum n+1 is the $LCM (1,2....9)=2520$ . So, the minimum $n=2519$ .