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## Number System: Basics of Factors Test-4

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Question 1 |

In the number 1600, how many factors are there which are not prefect squares?

12 | |

13 | |

11 | |

14 |

Question 1 Explanation:

Factors for 1600 are= 2

So the number of factors which are perfect square, 4x2 =8

How did we decide that? We know that for a factor to be a perfect square,

powers of prime numbers should be even, so number of ways that happens 2

5

So the number of factors in 1600 which are not a perfect square are 21-8=13

Option b is the right answer .

^{6}5^{2 }=(6+1)(2+1)=21So the number of factors which are perfect square, 4x2 =8

How did we decide that? We know that for a factor to be a perfect square,

powers of prime numbers should be even, so number of ways that happens 2

^{0}2^{2}2^{4}2^{6},5^{0},5

^{2}i.eÂ (3+1)(1+1)=8So the number of factors in 1600 which are not a perfect square are 21-8=13

Option b is the right answer .

Question 2 |

How many composite factors are there in 2

^{2}3^{3}5^{5}7^{7}11^{11}13^{13}?96768 | |

96775 | |

96761 | |

96765 |

Question 2 Explanation:

We have number N= 2

Number of factors are (2+1)(3+1)(5+1)(7+1)(11+1)(13+1)= 96768

Now there are 6 prime numbers (that is the six prime factors) are used and there is also 1,

1 is not a prime nor composite so the number of composite factors areÂ 96768-7=96761

^{2}3^{3}5^{5}7^{7}11^{11}13^{13Â }Number of factors are (2+1)(3+1)(5+1)(7+1)(11+1)(13+1)= 96768

Now there are 6 prime numbers (that is the six prime factors) are used and there is also 1,

1 is not a prime nor composite so the number of composite factors areÂ 96768-7=96761

Question 3 |

Calculate the sum of all even factors of 420.

576 | |

1056 | |

1150 | |

1152 |

Question 3 Explanation:

Step 1: For even factors divide the number with 2 thus 420/2=210

Step 2:Â 210= 2x3x5x7 (if we multiply each factor 210 by 2 it will give the even factor of 420),

so will find the sum of even factors of 210 and then we will multiply the result with 2

Step 3:Â {(2

(Use formula for sum of all factors)

Step 4: 576x2=1152

Hence option D is the right answer

Step 2:Â 210= 2x3x5x7 (if we multiply each factor 210 by 2 it will give the even factor of 420),

so will find the sum of even factors of 210 and then we will multiply the result with 2

Step 3:Â {(2

^{2}-1)/(2-1)x (3^{2}-1)/(3-1)x (5^{2}-1)/(5-1)x(7^{2}-1)/(7-1)}=576(Use formula for sum of all factors)

Step 4: 576x2=1152

Hence option D is the right answer

Question 4 |

If N=120, then how many co-primes are there which are less than the number N

32 | |

64 | |

18 | |

40 |

Question 4 Explanation:

Since this very simple problem if you know the formulae for this

Number of co- primes to the number N and also less than N is given by N (1-1/a)(1-1/b)(1-1/c)

Step 1:Â Prime factorisation for 120= 2

3

Step2:Â Â Â number of co-primes are = 120{(1-1/2)(1-1/3)(1-1/5)}=32.So option a is the right answer

Number of co- primes to the number N and also less than N is given by N (1-1/a)(1-1/b)(1-1/c)

Step 1:Â Prime factorisation for 120= 2

^{3}3

^{1}5^{1}Step2:Â Â Â number of co-primes are = 120{(1-1/2)(1-1/3)(1-1/5)}=32.So option a is the right answer

Question 5 |

How many pairs of factors are there for the number 900 such that they are co-prime to each other?

32 | |

64 | |

63 | |

128 |

Question 5 Explanation:

Remember these two formulas:

Number of pairs which are co-prime to each other is given by the formula for a number having two prime factors (p,q,r) :Â 1+(p+q)+2(pq)

Number of pairs which are co-prime to each other is given by the formula for a number having three prime factors (p,q,r) :Â 1+(p+q+r)+2(pq+pr+qr)+4pqr

So step1:Â 900 = 2

where p,q,r = 2,2,2, respectively.

Since 900 has 3 prime factors, we use the second formula.

Step 2: put the values in the above expression and get the value 63, so the right option is c

Number of pairs which are co-prime to each other is given by the formula for a number having two prime factors (p,q,r) :Â 1+(p+q)+2(pq)

Number of pairs which are co-prime to each other is given by the formula for a number having three prime factors (p,q,r) :Â 1+(p+q+r)+2(pq+pr+qr)+4pqr

So step1:Â 900 = 2

^{2}3^{2}5^{2 }where p,q,r = 2,2,2, respectively.

Since 900 has 3 prime factors, we use the second formula.

Step 2: put the values in the above expression and get the value 63, so the right option is c

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