Brian’s answer has near instantaneous evaluation in the Python shell when we enter:

>>> n = 2**105 * 3**140 * 5**84 * 7**90

>>> n

150462810922326152710290228433686961530697356776074449373600

141938371053848189980134027578261857302770024765419887333164

323078738017254430529707573248000000000000000000000000000000

000000000000000000000000000000000000000000000000000000

The number of digits in this large integer is also easily evaluated if we enter the following in the Python shell:

>>> print(len(str(n)))

234

We can also format the answer with thousands digit comma separators if we enter the following in the Python shell:

>>> print(“{:,d}”.format(n))

150,462,810,922,326,152,710,290,228,433,686,961,530,697,356,776,074,449,373,600,

141,938,371,053,848,189,980,134,027,578,261,857,302,770,024,765,419,887,333,164,

323,078,738,017,254,430,529,707,573,248,000,000,000,000,000,000,000,000,000,000,

000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

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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