Multiples and Factors MCQ Quiz - Objective Question with Answer for Multiples and Factors - Download Free PDF

Given:

Find smallest perfect square M divisible by 12, 15, and 18

Then divide M by 25 and find the sum of digits of quotient

Calculation:

Find LCM of 12, 15, 18

⇒ 12 = 2² × 3

⇒ 15 = 3 × 5

⇒ 18 = 2 × 3²

⇒ LCM = 2² × 3² × 5 = 180

⇒ 180 = 2² × 3² × 5

To make perfect square, multiply by 5 (missing square for 5)

⇒ M = 180 × 5 = 900

⇒ M / 25 = 900 / 25 = 36

Sum of digits of 36 = 3 + 6 = 9

∴ Required sum = 9

Given:

Number = 100

Formula Used:

The sum of all the factors of a^m × bⁿ is (a0 + a1 +......+ a^m) × (b0 + b1 +............+ bⁿ)

Calculation:

Factors of 100 = 2² × 5²

Sum of factors = (2⁰ + 2¹ + 2²) × (5⁰ + 5¹ + 5²)

⇒ (1 + 2 + 4) × (1 + 5 + 25)

⇒ 7 × 31

⇒ 217

∴ The sum of all factors of 100 is 217.

Formula Used:

A perfect cube has prime factors with exponents that are multiples of 3.

Calculation:

The prime factorization of 5673375 = 5³ × 3³ × 41².

To make this a perfect cube, the exponent of each prime factor must be a multiple of 3.

The prime factors and their current exponents are:

3: exponent 3 (already a multiple of 3)

5: exponent 3 (already a multiple of 3)

41: exponent 2 (needs one more factor to become a multiple of 3)

To make the exponent of 41 equal to 3, we need to multiply by 41^(3-2) = 41¹ = 41.

The smallest number to multiply 5673375 by to yield a perfect cube is 41.

Given:

Find the smallest number by which 675 must be multiplied to obtain a perfect cube?

Formula used:

Prime factorization of a number to find the smallest multiplier to make it a perfect cube.

Calculation:

Prime factorization of 675:

675 = 5 × 135

135 = 5 × 27

27 = 3 × 9

9 = 3 × 3

⇒ 675 = 5 × 5 × 3 × 3 × 3

To make it a perfect cube, each prime factor must appear in multiples of 3.

Here, 5 appears only twice, and it needs to appear three times.

⇒ The smallest number to multiply is 5

∴ The correct answer is option 2.

Given:

What is the e×ponent of 3 in the prime factorisation of 3825?

Formula used:

Prime factorisation involves e×pressing a number as a product of its prime factors.

Calculation:

3825 = 3 × 3 × 5 × 5 × 17

3825 = 3² × 5² × 17

Therefore, 3825 = 32 × 52 × 17

E×ponent of 3 = 2

∴ The correct answer is option (1).

Given:

3240

Concept:

If k = ax × by, then

a, and b must be prime number 

Sum of all factors = (a0 + a1 + a2 + ….. + ax) (b0 + b1 + b2 + ….. + by)

Solution:

3240 = 2³ × 3⁴ × 5¹

Sum of factors = (20 + 21 + 22 + 23) (3⁰ + 3¹ + 3² + 3³ + 3⁴) (5⁰ + 5¹)

⇒ (1 + 2 + 4 + 8) (1 + 3 + 9 + 27 + 81) (1 + 5)

⇒ 15 × 121 × 6

⇒ 10890

∴ required sum is 10890

Given:

The number is 8¹⁰ × 9⁷ × 7⁸ 

Concept used:

If a number of the form x^a × y^b × z^c ...... and so on, then total prime factors = a + b + c ..... and so on

Where x, y, z, ... are prime numbers

Calculation:

The number 8¹⁰ × 9⁷ × 7⁸ can be written as (2³)¹⁰ × (3²)⁷ × 7⁸ 

The number can ve written as 2³⁰ × 3¹⁴ × 7⁸ 

Total number of prime factors = 30 + 14 + 8

∴ The total number of prime factors are 52

Given:

Number = 100

Formula Used:

The sum of all the factors of a^m × bⁿ is (a0 + a1 +......+ a^m) × (b0 + b1 +............+ bⁿ)

Calculation:

Factors of 100 = 2² × 5²

Sum of factors = (2⁰ + 2¹ + 2²) × (5⁰ + 5¹ + 5²)

⇒ (1 + 2 + 4) × (1 + 5 + 25)

⇒ 7 × 31

⇒ 217

∴ The sum of all factors of 100 is 217.

Given:

The number of factors of 196 which are divisible by 4

Calculations:

According to the question,

The number 196 is exactly divisible = 1, 2, 4, 7, 14, 28, 49, 98, and 196.

Factors of 196 divisible by 4 = 4, 28, 196

∴ The number of factors of 196 which are divisible by 4 is 3.

Concept Used:

If N = a^p × b^q × c^r...

Then number of factors = (p + 1)(q + 1)(r + 1).....

Calculation:

According to the question,

First prime factorise number 720,

720 = 2⁴ × 3² × 5¹

On comparing, with concept,

p = 4, q = 2 and r = 1

No. of factors = (4 + 1)(2 + 1)(1 + 1)

⇒ 5 × 3 × 2 = 30

∴ The number of factors of 720 is 30.

Concept:

In this type of question, the square roots of factors will be natural number only, if the factors itself is a perfect square.

Calculation:

The number 2250 can be expressed as 2250 = 2 × 3² × 5³

Given:

The number is 480.

Formula Used:

If Prime Factorisation of a number N = a^p × b^q × c^r × ......

Then, Total number of factors of N = (p + 1) × (q + 1) × (r + 1) × ......

Calculation:

Here,

Prime Factorisation of 480 = 2⁵ × 3¹ × 5¹

So, Total number of factors = (5 + 1) × (1 + 1) × (1 + 1)

⇒ 6 × 2 × 2

⇒ 24

∴ The total number of factors of 480 is 24.

Calculations:

⇒ 213 + 214 + 215 + 216 + 217

⇒ 213 (1+ 21 + 22 + 23 + 24)

⇒ 213 (1 + 2 + 4 + 8 + 16)

⇒ 213 × (31)

∴ 213 + 214 + 215 + 216 + 217 is a multiple of 31.

Concept Used:

Number = a^p × b^q × c^r...

Total number of factors = (p + 1)(q + 1)(r + 1) ...

Total number of odd factors = (q + 1)(r + 1) ..., if 'a' is an even prime number

Here, a, b, c... etc., are prime numbers.

Calculation:

Prime factorization of 378:

378 = 2 × 3 × 3 × 3 × 7

378 = 2¹ × 3³ × 7¹

So, the total number of factors of 378

⇒ (1 + 1)(3 + 1)(1 + 1)

⇒ 2 × 4 × 2 = 16

Conclusion:

The number 378 can be divided equally among the students in 16 different ways, as it has 16 factors. So, your solution is correct.

Therefore, the required number of ways = 16.

Concept Used: 

Each integer can be expressed by multiple of several prime numbers or polynomials of those prime numbers.  

Formula Used: 

If a number N can be expressed as

N = am ×  bⁿ × c^p (where a, b, and c are prime numbers)

Then total number of factors is equal to (m + 1) × (n + 1) × (p + 1) 

Calculation: 

Factors of 540 are numbers that, when multiplied in pairs give the product as 540.

There are 24 factors of 540, of which the following are its prime factors 2, 3, 5.

The Prime factorization of 540 is 2² × 3³ × 5¹.

Here, m = 2, n = 3 , p = 1

The total number of  factors of 540 are: 

⇒ (2 + 1) × (3 + 1) × (1 + 1) = 24

So, Total number of factors of 540 is = 24

Hence, the correct answer is "24".