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8 years, 8 months ago
8 years, 7 months ago
The number of 7s that 42! has is [$$\frac{42}{7}$$] = 6
The number of 7s that 56! has is [$$\frac{56}{7}$$] + [$$\frac{56}{7^2}$$] = 9
So, all the values from 42! to 55! are multiples of $$7^6$$
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