What is the smallest n that fits the requirements?

Question

What is the smallest value of $n$ if $n(n+1)(n+2)(n+3) \equiv 0 \pmod{2000}$

Answer 1

Observe here that $5$ will divide atmost one term out of $n,n+1,n+2,n+3$ because these are four consecutive natural numbers (assuming $n>0$). Therefore $125$ will be the factor of one these term. Also $2$ will always divide two term in these and $4$ will divide atmost one term , this means we need not to care about factor $8$ but we want little more $i.e$ some term which is divisible by $8$ not just $4$. Now hoping one term to be $125$ we look upon product of four possible consecutive numbers out of $122,123,124,125,126,127,128$ we found that $128\equiv0($ mod $16)$. Therefore we take $125,126,127,128$. Consequently , $$125\times126\times127\times128\equiv0(mod \quad2000)$$ And hence smallest $n$ is $125$.