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

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    Does $n=0$ count? What about $n=-3$? – 2017-01-21
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    I suppose what I was trying to ask is this: *Are we restricting ourselves to the positive integers?* – 2017-01-21
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    Assuming you meant $n>0$, Hint: at most one of those terms can be divisible by $5$ hence that term must be divisible by $5^3$. – 2017-01-21
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    I'm sorry for not adding, **n** is a natural number. And thanks for the hint, i figured it out. – 2017-01-21
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    Also at most one term will be divisible by four. Another by 2 but not 4. 2000 =125 x 16 so the term divisible by 4 must be divisible by 8. So you need the smallest multiple of 125 that is 3 with a multiple of 8. – 2017-01-21
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    If you have figured it out, John, let me encourage you to post an answer. Then, you can accept it. – 2017-01-21

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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 $125$.