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## Number System: Basics of Factors Test-1

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Question 1 |

How many factors does 825 have

14 | |

16 | |

12 | |

20 |

Question 1 Explanation:

**Step 1:**Prime factorisation, so N=825= 3

^{1}5

^{2}11

^{1}

Power of 3 as 3

^{0 }, 3

^{1}( 1+1=2)ways ,

Power of 5 as 5

^{0 }, 5

^{1},5

^{2 }( 2+1=3)ways

Power of 11 as 11

^{0}, 11

^{1 }(1+1=2)ways

**Step 2:**Hence, the number of factors is (1+1)(2+1)(1+1)=2x3x2=12

Question 2 |

How many are number of factors N= 3

^{2}5^{3}7^{3}11^{2}13^{2}17 ?700 | |

456 | |

864 | |

900 |

Question 2 Explanation:

**Step 1:**Prime factorisation, so N=3

^{2}5

^{3}7

^{3}11

^{2}13

^{2}17

^{ 1}

Power of 3 as 3

^{0 }, 3

^{1},3

^{2 }

Power of 5 as 5

^{0 }, 5

^{1},5

^{2},5

^{3}

Power of 7 as 7

^{0},7

^{1},7

^{2},7

^{3}

Power of 11as 11

^{0}11

^{1}11

^{2}

Power of 13 as 13

^{0}13

^{1}13

^{2}

Power of 17 as 17

^{0}17

^{1}

**Hence, the number of factors is(2+1)(3+1)(3+1)(2+1)(2+1)(1+1)=864**

Step 2:

Step 2:

Question 3 |

How many even factors does the number 2

^{4 }3^{5}7^{2 }have?76 | |

72 | |

36 | |

128 |

Question 3 Explanation:

In this case we have to find number of even factors,an even factor is divisible by 2 or smallest power of 2 has to be 1 not 0.

Hence a factor must be 2

Hence total number of factors= (4)(5 + 1 )(2 + 1) = 72

Hence a factor must be 2

^{(1 or 2 or 3 or 4)}3^{(0 or 1,2,3,4,5)}7^{(0 or 1 or 2) }Hence total number of factors= (4)(5 + 1 )(2 + 1) = 72

Question 4 |

How many odd factors does the number N= 2

^{4 }3^{5}7^{2}?20 | |

40 | |

45 | |

18 |

Question 4 Explanation:

To simply find odd factors, remove 2 from prime factors i.e from N remove ${{2}^{4}}$

Then we are left with $N={{3}^{5}}{{7}^{2}}$ Since these are two odd factors, they will have no even factors.

Therefore, the total number of factors is= (5+1)(2+1)=18

Then we are left with $N={{3}^{5}}{{7}^{2}}$ Since these are two odd factors, they will have no even factors.

Therefore, the total number of factors is= (5+1)(2+1)=18

Question 5 |

How many factors are there in N =

**2**^{5 }5^{2}7^{2}which are divisible by 10?45 | |

35 | |

30 | |

55 |

Question 5 Explanation:

If a number is divisible by 10 then it must have minimum power of 2 and 5 as 1.

A factor divisible by 10 is 2

Hence, total number of factors divisible by 10 is = (5)(2)(2 + 1) = 30

A factor divisible by 10 is 2

^{(1 or 2 or 3or 4 or 5)}5^{(1 or 2 )}7^{(0 or 1 or 2 )}Hence, total number of factors divisible by 10 is = (5)(2)(2 + 1) = 30

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