The number of distinct factors a number has is given by the product of one
more than the maximum power of each of the prime factors.
the number 24*33 = 792 can be written as (2^3)*(3^2)*(11^1)
then the number of distinct factors of 792 will become (3+1)*(2+1)*(1+1) =
24
So we can form 24/2 = 12 distinct sets of (a,b)  such that N = a*b

http://www.mathisfunforum.com/viewtopic.php?id=11042


On Tue, Aug 23, 2011 at 11:38 PM, Shravan Kumar <[email protected]> wrote:

> It would be sum of number of  factors of both the numbers.
> 24 -1,2,3,4,6,8,12,24
> 33-1,3,11,33
>
>
> On Tue, Aug 23, 2011 at 10:54 PM, Aman Goyal <[email protected]>wrote:
>
>> ans is 12, but instead of counting i am looking for some better solution.
>>
>>
>> On Tue, Aug 23, 2011 at 10:48 PM, manish patel 
>> <[email protected]>wrote:
>>
>>>
>>> (24,33),(12,66),(8,99),(6,132),(4,198),(3,254),(2,396),(1,792),(792,1),(72,11),(264,3),(33,24)
>>>
>>> On Tue, Aug 23, 2011 at 10:18 PM, Aman Goyal <[email protected]>wrote:
>>>
>>>> Let a natural number N be such that N = a × b where a and b are the
>>>> factors of N. How many such sets of (a, b) can be formed in which the
>>>> selection of the two numbers a and b is distinctly different if N = 24 × 
>>>> 33?
>>>>
>>>> Please explain your solution also.
>>>>
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>>>
>>>
>>>
>>> --
>>> With Regards
>>>
>>> Manish Patel
>>> BTech 3rd Year
>>> Computer Science And Engineering
>>> National Institute of Technology -Allahabad
>>>
>>>
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