Find the number of positive divisors of 1 0 ! .
Notation : ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 .
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this problem made made me write a piece of code for this program in c
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Ahahah. I've just done the same, but in pascal😂
Follow up: How many primes are contained in 10! ?
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obviously 4 [2,3,5,7]
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What is that supposed to mean? I mean , I don't know the syntax of the operator you mentioned
The prime divisor numbers of n! equals to [n/p]+[n/p^2]+[n/p^3]...... The prime numbers less than 10 are 2,3,5and7. 2: 5+2+1+0+0+........=8 3:3+1=4 5: 2 7: 1 All the divisors are the combine of these primes. Use the fomula: number=(8+1)(4+1)(2+1)(1+1)=270
Where this formula came from?
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You mean the number of divisors after knowing the number of each prime divisor?
You can visualise the formula like this: Suppose we have 3 6 0 = 2 3 . 3 2 . 5 1
While counting the number of divisors, we have 4 options to choose power of 2 (viz 0, 1, 2 or 3)
We have 3 options to choose power of 3 (viz 0, 1 or 2)
We have 2 options to choose power of 5 (0 or 1)
So in total we have 4 x 3 x 2 = 24 divisors to choose from (this will include 1 when we choose all bases with 0 power; and 360 when we choose all bases with their full powers)
In general, N = 2 a . 3 b . 5 c . . . . . . . will have (a+1)(b+1)(c+1)...... divisors, including 1 and N itself.
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1 0 ! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 1 0 = 2 8 × 3 4 × 5 2 × 7 1 .
∴ Number of divisors = 9 × 5 × 3 × 2 = 2 7 0 .
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