Similar arguments for 3, 5 and 7 lead to the form of number as:\[\begin{align}N&=2^{3\cdot 5\cdot 7\cdot a} \cdot 3^{2\cdot 5\cdot 7\cdot b} \cdot 5^{2\cdot 3\cdot 7\cdot c} \cdot 7^{2\cdot 3\cdot 5\cdot d}\\&= 2^{105a} \cdot 3^{70b} \cdot 5^{42c} \cdot 7^{30d}\end{align}\]

for integer \(a\), \(b\), \(c\) and \(d\). Now:

1) \(2N\) is a square which means that:\[105a+1=0\mod2\implies a=1\mod2\]

2) \(3N\) is a cube which means that:\[70b+1=0\mod3\implies b=2\mod3\]

3) \(5N\) is a fifth power which means that:\[42c+1=0\mod5\implies c=2\mod5\]

4) \(7N\) is a seventh power which means that:\[30d+1=0\mod7\implies c=3\mod7\]Since we need the smallest value of \(N\), we hence obtain:\[N=2^{105}\cdot3^{140}\cdot5^{84}\cdot7^{90}\]

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