Find the number of positive integers n<1000n < 1000n<1000 such that GCD(n,n+1)+GCD(n,n+2)+GCD(n,n+3)=10\text{GCD}(n, n+1) + \text{GCD}(n, n+2) + \text{GCD}(n, n+3) = 10GCD(n,n+1)+GCD(n,n+2)+GCD(n,n+3)=10.
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