Number System: Basics of Factors Test-2

The number of factors are given by the formula (p+1)(q+1)(r+1), where p,q,r are the powers of the prime factors of the number.
Step 1: prime factorisation
i.ea^pb^qc^r form
i.e  2²3²5¹ Step 2: (2+1)(2+1)(1+1) = 18
So the option c is the right answer

In the problem we are asked to find the factors having three digits, so insteadof the long procedure, we adopt a short-cut.
Simply divide 220 by 2, i.e220/2 =110 110/2=55
now 55 is two digit number, hence we can’t include it in the answer.
All subsequent factors would be less than 110.
So the total number of factors which are three digit numbers are 2 i.e. 220,110.
So option 1 is the right answer

\[\begin{align} & Product\text{ }of\text{ }co-prime\text{ }numbers\text{ }can\text{ }be\text{ }identified\text{ }by\text{ }{{2}^{k}}-1\text{ } \\ & where\text{ }k\text{ }is\text{ }the\text{ }number\text{ }of\text{ }prime\text{ }numbers\text{ } \\ & used\text{ }in\text{ }the\text{ }prime\text{ }factorisation. \\ \end{align}\]
\[Therefore:\] \[Step\text{ }1:\text{ }Prime\text{ }factorisation\text{ }75\text{ }=\text{ }{{5}^{2}}{{3}^{1}}\] \[Step\,\,2\text{ }:\text{ }There\text{ }are\text{ }two\text{ }prime\text{ }numbers\text{ }are\text{ }used\text{ }\] \[so\text{ }{{2}^{2}}-1=\text{ }3\text{ }so\text{ }the\text{ }right\text{ }answer\text{ }is\text{ }option\text{ }1.\] \[The\text{ }2\text{ }products\text{ }of\text{ }co-prime\text{ }numbers\text{ }in\text{ }this\text{ }case\text{ }\] \[are\text{ }1\times 75\text{ }and\text{ }3\times 25.\]

Product of divisors means the product of all the factors of number ,
the product of all the factors may be calculated as N^X/2
 where N is the number and X is  number of factors ,
so the product of all the factors or product of divisors is given by option B,
that is (2³3¹5¹)⁸

Step 1: 36 = 2²3²
Step 2: Number of factors (2+1)(2+1)=9
Step 3: Product of two distinct factors= (Number of factors + 1)/2 ; if the number is a perfect square
Number of ways to express a number as a product of two factors, = (Number of factors)/2 ;
if the number is not a perfect square Illustration of the case above:
Take the number 6 Number of factors = 4= 1,2,3,6
can be written as: 1*6 , 2*3
We can’t repeat 6*1 and 3*2 as they are already taken. Hence only 2 ways.
Another example: take 9.
Number of factors = 3 =1,3,9
Can be written as 1*9 only.  Hence only 1 way.
Hence number of ways = (9-1)/2=4 so the right option is d
These four pairs will be: (2,18), (1, 36), (4, 9) and (3, 12).
Remember in the problem we are asked the distinct factors so we are not using 6x6, in which both the numbers are same.