Find the smallest natural number `n` such that n! is divisible by 990.
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Answered by
1
Consider what the prime factors are that are used to produce 990. All the factors other than 11 can be found as common factors in factorial less than 11!, but 11 itself only is included in factorials greater than 10!. So n must be 11, and thus n! would be 11!.
Answered by
3
Answer:
smallest natural number is 1
Step-by-step explanation:
990÷1= 990
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