if my answer is correct then ok..but if it is not correct, please dont tell answer now, i will try once more then i will ask you for the answer..

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+2 votes

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is it 135 ?

if my answer is correct then ok..but if it is not correct, please dont tell answer now, i will try once more then i will ask you for the answer..

if my answer is correct then ok..but if it is not correct, please dont tell answer now, i will try once more then i will ask you for the answer..

+1

**Case 1: Non square no:**

Let’s take a simple no: 84.

Factors of 84 are:

1,2,3,4,6,7,12,14,21,28,42,84.

Total: 12 Let’s check if we miss any factor:

Total no of factors for 84: [(2^2)*3*7] = 3*2*2=12.

So, we captured all the factors of 84.

Now 84 can be expressed as a product of 2 natural no’s:

1*84, 2*42, 3*28, 4*21, 6*14, 7*12 ------- 6 Ways So if a no has n factors, it can be expressed as a product of 2 natural no’s as:

---- {(First factor from left)*(first factor from right)} ,

{(second factor from left)*(second factor from right)} ,

till………………………. {([n/2]nd factor from left) * ([n/2]nd factor from right)}.

**So, if a no has ‘N’ factors, where N is even (for non square number), total no of ways it can be expressed as product of 2 natural no’s: N/2.**

If you are clear till this point, you are thinking about what if a no has odd no of factors.

This leads to our case no 2.

**Case 2: Square number:** As many of you might be aware that **Only **a square number has odd no of factors. Or if a no has odd no of factors, then it has to be a square no.

lets take a simple no: 36

Factors of 36: 1,2,3,4,6,9,12,18,36.

Total no of factors: 9.

36 can be expressed as a product of 2 different natural no’s:

1*36, 2*18, 3*12, 4*9, 6*6. ------ 5 ways.

If you notice, middle no is expressed as a product to itself (6*6).

**So, for a square no having no of factors as N, total no of ways are:**

**(N+1)/2 ------** **If repetition of numbers is allowed.**

**(N-1)/2-------** **If both the no’s has to be unique.**

---------------------------------------------------------------

+4 votes

Best answer

Lets try to prime factorize $10!$ because each if $p^n$ is a factor of a number $x$ for any prime number $p$, then $x$ has at least $p+1$ factors -- $p^0, p^1, p^2, \ldots, p^n.$

$10!=2\times3\times4\times5\times6\times7\times8\times9\times10.$

$=2^8 \times 3^4 \times 5^2 \times 7$

Now, we can count the no. of factors of $10!$ which will be $(8+1) \times (4+1) \times (2+1) \times (1+1) =270.$ For each of the factor $d$ of any number $x$, we can get $x$ by multiplying $d$ by $x/d$ where $x/d$ is another factor of $x$. So, number of ways we can choose $(a,b)$ such that $a\times b = 10!$ will be $\frac{270}{2} = 135.$ In multiplication there is no need to consider $a \times b$ different from $b \times a$, otherwise answer would be $270$.

$10!=2\times3\times4\times5\times6\times7\times8\times9\times10.$

$=2^8 \times 3^4 \times 5^2 \times 7$

Now, we can count the no. of factors of $10!$ which will be $(8+1) \times (4+1) \times (2+1) \times (1+1) =270.$ For each of the factor $d$ of any number $x$, we can get $x$ by multiplying $d$ by $x/d$ where $x/d$ is another factor of $x$. So, number of ways we can choose $(a,b)$ such that $a\times b = 10!$ will be $\frac{270}{2} = 135.$ In multiplication there is no need to consider $a \times b$ different from $b \times a$, otherwise answer would be $270$.

0 votes

We know that 10! = 3628800

Prime Factors for the number 3628800 = (2^{8} × 3^{4} × 5^{2} × 7) or(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 7) = 269 factor

** If a Number has ‘N’ factors, Where N is odd (for non square number), Total number of ways it can be expressed as product of 2 natural no’s:**

**(N+1)/2 ------** **If repetition of numbers is allowed.**

**(N-1)/2-------** **If both the no’s has to be unique.** ** **

** ways can 10! be written as the product of two natural number = Total factor / 2 = 269+1 / 2 = 135 ans**

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