If \(x\) and \(n\) are integers then \[(1-x)^n (2-x)^{2n} (3-x)^{3n} (4-x)^{4n} (5-x)^{5n}\] is

negative when \(n>5\) and \(x<5\),

negative when \(n\) is odd and \(x>5\),

negative when \(n\) is a multiple of \(3\) and \(x>5\),

negative when \(n\) is even and \(x<5\).

For convenience, let \[f_n(x) = (1-x)^n (2-x)^{2n} (3-x)^{3n} (4-x)^{4n} (5-x)^{5n}.\]

First consider the case \(x=4, n = 6\). We have \(f_6(4)=0\), which means that (a) and (d) are false.

Now consider the case \(x=6, n=6\). Then each of \((1-x)^n\), \((2-x)^{2n}\), \((3-x)^{3n}\), \((4-x)^{4n}\) and \((5-x)^{5n}\) is greater than or equal to zero, which means that (c) is false.

It only remains to show that (b) is true. If \(x>5\), then each of \(1-x\), \(2-x\), \(3-x\), \(4-x\) and \(5-x\) is negative.

If \(n\) is odd here then \[\begin{align*} (1-x)^n &<0\\ (2-x)^{2n}&>0\\ (3-x)^{3n}&<0\\ (4-x)^{4n}&>0\\ (5-x)^{5n}&<0 \end{align*}\]so their product is negative, as required. The answer is (b).