find smallest value of N such that N mod 10 = 9 , N mod 9 = 8 and so on

Question

find the smallest value of N such that N mod 10=9, N mod 9 = 8, N mod 8 = 7, N mod 7 = 6 and so on till N mod 2 = 1 .

Answer 1

Hint $\,\ 2,3,\cdots, 10\mid N\!+\!1 \iff {\rm lcm}(2,3,\cdots,10)\mid N\!+\!1$

Answer 2

Add to both sides of congruents $+1$. We know that $\text{lcm}(2,3,4,5,6,7,8,9,10)=2520$ and we get $N+1\equiv 0 \pmod {2520}$. Minimal value of $N$ will be $2519$