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## Number System: Basics of Factors Test-2

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Question 1 |

Find the number of factors 180

32 | |

20 | |

18 | |

22 |

Question 1 Explanation:

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

i.eÂ 2

So the option c is the right answer

Step 1: prime factorisation

i.ea

^{p}b^{q}c^{r}formi.eÂ 2

^{2}3^{2}5^{1}Step 2: (2+1)(2+1)(1+1) = 18So the option c is the right answer

Question 2 |

How many factors of 220 are three digit numbers?

2 | |

4 | |

1 | |

3 |

Question 2 Explanation:

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 canâ€™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

Simply divide 220 by 2, i.e220/2 =110 110/2=55

now 55 is two digit number, hence we canâ€™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

Question 3 |

In how many ways can 75 be written as the product of its co-prime numbers?

3 | |

5 | |

6 | |

1 |

Question 3 Explanation:

\[\begin{align}
& Product\text{ }of\text{ }co-prime\text{ }numbers\text{ }can\text{ }be\text{ }identified\text{ }by\text{ }{{2}^{k}}-1\text{ } \\
& where\text{ }k\text{ }is\text{ }the\text{ }number\text{ }of\text{ }prime\text{ }numbers\text{ } \\
& used\text{ }in\text{ }the\text{ }prime\text{ }factorisation. \\
\end{align}\]

\[Therefore:\] \[Step\text{ }1:\text{ }Prime\text{ }factorisation\text{ }75\text{ }=\text{ }{{5}^{2}}{{3}^{1}}\] \[Step\,\,2\text{ }:\text{ }There\text{ }are\text{ }two\text{ }prime\text{ }numbers\text{ }are\text{ }used\text{ }\] \[so\text{ }{{2}^{2}}-1=\text{ }3\text{ }so\text{ }the\text{ }right\text{ }answer\text{ }is\text{ }option\text{ }1.\] \[The\text{ }2\text{ }products\text{ }of\text{ }co-prime\text{ }numbers\text{ }in\text{ }this\text{ }case\text{ }\] \[are\text{ }1\times 75\text{ }and\text{ }3\times 25.\]

\[Therefore:\] \[Step\text{ }1:\text{ }Prime\text{ }factorisation\text{ }75\text{ }=\text{ }{{5}^{2}}{{3}^{1}}\] \[Step\,\,2\text{ }:\text{ }There\text{ }are\text{ }two\text{ }prime\text{ }numbers\text{ }are\text{ }used\text{ }\] \[so\text{ }{{2}^{2}}-1=\text{ }3\text{ }so\text{ }the\text{ }right\text{ }answer\text{ }is\text{ }option\text{ }1.\] \[The\text{ }2\text{ }products\text{ }of\text{ }co-prime\text{ }numbers\text{ }in\text{ }this\text{ }case\text{ }\] \[are\text{ }1\times 75\text{ }and\text{ }3\times 25.\]

Question 4 |

TheÂ product of divisors ofÂ 120 will be

(2 ^{3}3^{1}5^{1})^{16} | |

(2 ^{3}3^{1}5^{1})^{8} | |

(2 ^{3}3^{1}5^{1})^{20} | |

(2 ^{3}3^{1}5^{1})^{4} |

Question 4 Explanation:

Product of divisors means the product of all the factors of number ,

the product of all the factors may be calculated as N

Â where N is the number and X isÂ number of factors ,

so the product of all the factors or product of divisors is given by option B,

that is (2

the product of all the factors may be calculated as N

^{X/2 }Â where N is the number and X isÂ number of factors ,

so the product of all the factors or product of divisors is given by option B,

that is (2

^{3}3^{1}5^{1})^{8}Question 5 |

In how many number of ways 36 can be expressed as the product of two distinct factors?

10 | |

5 | |

8 | |

4 |

Question 5 Explanation:

Step 1: 36 = 2

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,Â = (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*6Â , 2*3

We canâ€™t repeat 6*1 and 3*2 as they are already taken.Â Hence only 2 ways.

Another example: take 9.

Number of factors = 3Â =1,3,9

Can be written as 1*9 only.Â Hence only 1 way.

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.

^{2}3^{2}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,Â = (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*6Â , 2*3

We canâ€™t repeat 6*1 and 3*2 as they are already taken.Â Hence only 2 ways.

Another example: take 9.

Number of factors = 3Â =1,3,9

Can be written as 1*9 only.Â 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.

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