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Number System: Basics of Factors Test-3
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Question 1 |
How many factors are there in N = 25 55 72112 which ends with zero?
230 | |
225 | |
105 | |
270 |
Question 1 Explanation:
This question talks about the factors that end with zero.
All we need to find is number of factors that end with zero.
For this, the minimum power of 2 and 5 has to be one.
A factor divisible by 10 is of the form 2(1 or 2 or 3or 4 or 5)5(1 or 2 or 3 or4 or5 Â )7(0 or 1 or 2 )11(0,1,2)
Hence, total number of factors divisible by 10 is = (5)(5)(2 + 1)(2+1) = 225
All we need to find is number of factors that end with zero.
For this, the minimum power of 2 and 5 has to be one.
A factor divisible by 10 is of the form 2(1 or 2 or 3or 4 or 5)5(1 or 2 or 3 or4 or5 Â )7(0 or 1 or 2 )11(0,1,2)
Hence, total number of factors divisible by 10 is = (5)(5)(2 + 1)(2+1) = 225
Question 2 |
How many number of factors are in N = 23 325272 which is divisible by 20?
72 | |
70 | |
38 | |
36 |
Question 2 Explanation:
Since we have to find the number of factors which are divisible by 20,
to find this divide the given number by 20 and then find the number of factors of the quotient.
Divide 23325272 by 20 ( 2251)
= 21 32 51 72 and its number of factors are 2x3x2x3=36
to find this divide the given number by 20 and then find the number of factors of the quotient.
Divide 23325272 by 20 ( 2251)
= 21 32 51 72 and its number of factors are 2x3x2x3=36
Question 3 |
How many number of factors of N= 24 33112133 that are perfect square?
20 | |
22 | |
24 | |
30 |
Question 3 Explanation:
We know that for a number to be a perfect square, its factor must have the even number of powers.
All we need to explore is the even powers of the factors in this case.
Therefore the perfect square factors can be2 ( o or 2 or 4Â Â )3(0 or 2 )11(0 or 2 )13(0 or 2)
Hence, the total number of factors which are perfect square is 3x2x2x2=24
All we need to explore is the even powers of the factors in this case.
Therefore the perfect square factors can be2 ( o or 2 or 4Â Â )3(0 or 2 )11(0 or 2 )13(0 or 2)
Hence, the total number of factors which are perfect square is 3x2x2x2=24
Question 4 |
How many factors of N= 2936118 that are perfect cube?
72 | |
36 | |
45 | |
20 |
Question 4 Explanation:
If a number is a perfect cube, then the power of the prime factors should be divisible by 3.
The possible options are : 2(0 or 3 or 6 or 9)3(0 or 3 or 6)11 (0 or 3 or 6 ) .
 Hence the total number of factors which are perfect cube 4 x 3x3 =36.
The possible options are : 2(0 or 3 or 6 or 9)3(0 or 3 or 6)11 (0 or 3 or 6 ) .
 Hence the total number of factors which are perfect cube 4 x 3x3 =36.
Question 5 |
How many factors of N= 2936118are both perfect square and perfect cube?
8 | |
12 | |
14 | |
10 |
Question 5 Explanation:
If a number is perfect square and perfect cube then the powers of prime factors must be divisible by 6.
Perfect square factor must be 2(0 or 6 or 9)3(0 or 6)7(0 or 6)
Hence total number of perfect cube factors is 3 x 2 x 2 = 12
Perfect square factor must be 2(0 or 6 or 9)3(0 or 6)7(0 or 6)
Hence total number of perfect cube factors is 3 x 2 x 2 = 12
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