How many values can natural number n take, if n! Is multiple or 7^6 but not of 7^9 ? approach plz

Question

Answer 1

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 $$7^6$$

Answer 2

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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How many values can natural number n take, if n! Is multiple or 7^6 but not of 7^9 ? approach plz