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in how many ways can 10! be written as the product of two natural number?
in Numerical Ability by Active | 166 views
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..
incorrect ans
270 ??

but i have taken 2*4 and 4*2 two different entities..
exactly u got it

In how many ways can a number be written as a product of two different factors?

Case 1: Non square no:

Let’s take a simple no: 84.
Factors of 84 are:
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.

2 Answers

+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 -- $p^0, p^1, p^2, \ldots, p^n.$


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

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 $270$.
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 We know that    10! = 3628800 

Prime Factors for the number 3628800 = (28 × 34 × 52 × 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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