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Find the number of positive divisors of 10 ! 10! .

Notation : ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × × 8 8! = 1\times2\times3\times\cdots\times8 .


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The answer is 270.

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2 solutions

Swapnil Das
Jun 14, 2016

10 ! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 = 2 8 × 3 4 × 5 2 × 7 1 10!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10={ 2 }^{ 8 }\times { 3 }^{ 4 }\times { 5 }^{ 2 }\times { 7 }^{ 1 } .

\therefore Number of divisors \text{ Number of divisors} = 9 × 5 × 3 × 2 = 270 = 9\times 5\times 3\times 2=270 .


Relevant article

this problem made made me write a piece of code for this program in c

Shashank Rustagi - 5 years ago

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Ahahah. I've just done the same, but in pascal😂

Alex Misiurina - 4 years, 12 months ago

Follow up: How many primes are contained in 10! ?

Mehul Arora - 4 years, 12 months ago

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obviously 4 [2,3,5,7]

fahim saikat - 4 years, 12 months ago

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What is that supposed to mean? I mean , I don't know the syntax of the operator you mentioned

Mehul Arora - 4 years, 12 months ago
Refb Pe
Jun 15, 2016

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?

Daniel Brazil - 4 years, 12 months ago

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You mean the number of divisors after knowing the number of each prime divisor?

Refb Pe - 4 years, 12 months ago

You can visualise the formula like this: Suppose we have 360 = 2 3 . 3 2 . 5 1 360 = 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 . . . . . . . N = 2^{a}.3^{b}.5^{c}....... will have (a+1)(b+1)(c+1)...... divisors, including 1 and N itself.

Rohit Sachdeva - 4 years, 12 months ago

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