Divisibility and Remainder MCQ Quiz - Objective Question with Answer for Divisibility and Remainder - Download Free PDF

Last updated on Jul 2, 2025

To master your preparation for divisibility and remainder solve these Divisibility and Remainder questions by Testbook. These questions are of an intermediate level and come with answers followed by the in depth explanation. In addition, to ease your preparation, a few tips and tricks are also stated. Divisibility and Remainder MCQ Quiz are formatted in a way that it’ll help you to ace up your preparation for this section. Begin your preparation by solving these questions.

Latest Divisibility and Remainder MCQ Objective Questions

Divisibility and Remainder Question 1:

The smallest 1-digit number to be added to the 6-digit number 348510 so that it is completely divisible by 11 is:

  1. 3
  2. 7
  3. 9
  4. 5
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 3

Divisibility and Remainder Question 1 Detailed Solution

Given:

The 6-digit number is 348510

Formula used:

To check divisibility by 11, the difference between the sum of the digits in odd positions and the sum of the digits in even positions should be either 0 or a multiple of 11.

Calculation:

Sum of digits in odd positions (3, 8, 1):

3 + 8 + 1 = 12

Sum of digits in even positions (4, 5, 0):

4 + 5 + 0 = 9

Difference between these sums:

12 - 9 = 3

To make 348510 divisible by 11, the difference should be a multiple of 11:

⇒ 3 + x = 0 or 11

⇒ x = 8 or x = 3

Therefore, the smallest 1-digit number to be added is 3

∴ The correct answer is option (1).

Divisibility and Remainder Question 2:

The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as remainder. What is the smaller number?

  1. 240
  2. 270
  3. 295
  4. 360
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 270

Divisibility and Remainder Question 2 Detailed Solution

Given:

The difference of two numbers = 1365

Quotient when the larger number is divided by the smaller number = 6

Remainder when the larger number is divided by the smaller number = 15

Formula Used:

Larger number = Smaller number × Quotient + Remainder

Difference = Larger number - Smaller number

Calculation:

Let the smaller number be x.

Larger number = 6x + 15

Given, difference = 1365

⇒ 6x + 15 - x = 1365

⇒ 5x + 15 = 1365

⇒ 5x = 1365 - 15

⇒ 5x = 1350

⇒ x = 1350 / 5

⇒ x = 270

The smaller number is 270.

Divisibility and Remainder Question 3:

If adding the three-digit number 4x3 to another three-digit number 984 gives a four-digit number 13y7 which is divisible by 11. then (x + y) = ?

  1. 10
  2. 11
  3. 12
  4. 15
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 10

Divisibility and Remainder Question 3 Detailed Solution

Concept used:

Divisibility Rules for 11:

If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.

Calculation:

13y7 is divisible by 11,

So, 13y7 = 1 + y = 3 + 7

⇒ y = 10 - 1 = 9

The number is 1397.

According to the question,

4x3 + 984 = 1397

⇒ 4x3 = 1397 - 984

⇒ 4x3 = 413

⇒ x = 1

Now, (x + y) = 1 + 9 = 10

∴ (x + y) = 10

Divisibility and Remainder Question 4:

If an 8-digit number 256139A4 is divisible by 11, find the value of A.

  1. 9
  2. 8
  3. 6
  4. 7
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 8

Divisibility and Remainder Question 4 Detailed Solution

Given:

The 8 digit number = 256139A4

Concept used:

Divisibility rule of 11 - Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.

Calculation:

According to the question

The 8 digit number = 256139A4

⇒ (2 + 6 + 3 + A) = (5 + 1 + 9 + 4)

⇒ 11 + A = 19

⇒ A = (19 – 11)

⇒ A = 8

∴ The required value of A is 8

Divisibility and Remainder Question 5:

Find the sum of all 3 digit numbers that leave a remainder of 5 when divided by 10. 

  1. 72000
  2. 54800
  3. 56500
  4. 49500
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : 49500

Divisibility and Remainder Question 5 Detailed Solution

Given:

Divisor = 10

Remainder = 5

Formula Used:

Last term, l = a + (n – 1)d

Sum of n terms = [2a + (n – 1)d]

Calculation:

First number, a = 105

Last number, l = 995

Common difference, d = 10

 According to AP last term formula,

⇒ 995 = 105 + (n – 1) × 10

⇒ 995 – 105 = 10n – 10

⇒ 890 + 10 = 10n

⇒ n =  = 90

Sum of all terms = [2 × 105 + (90 – 1) × 10]

Sum of all terms = 45 × (210 + 890)

Sum of all terms = 45 × 1100 = 49500

Therefore, the required sum of all 3-digit numbers is 49,500.

Top Divisibility and Remainder MCQ Objective Questions

Which of the following numbers is a divisor of ?

  1. 46
  2. 14
  3. 8
  4. 50

Answer (Detailed Solution Below)

Option 3 : 8

Divisibility and Remainder Question 6 Detailed Solution

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Given:

Concept used:

an​​​​​​ - bn is divisible by (a + b) when n is an even positive integer.

Here, a & b should be prime number.

Calculation:

⇒ 

⇒ 

Here, 30 is a positive integer.

​According to the concept,

 is divisible by (7 + 1) i.e., 8.

∴ 8 is a divisor of .

If the 5-digit number 676xy is divisible by 3, 7 and 11, then what is the value of (3x - 5y)?

  1. 9
  2. 11
  3. 10
  4. 7

Answer (Detailed Solution Below)

Option 1 : 9

Divisibility and Remainder Question 7 Detailed Solution

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Given:

676xy is divisible by 3, 7 & 11

Concept:

When 676xy is divisible by 3, 7 &11, it will also be divisible by the LCM of 3, 7 &11. 

Dividend = Divisor × Quotient + Remainder

Calculation:

LCM (3, 7, 11) = 231

By taking the largest 5-digit number 67699 and divide it by 231.

∵ 67699 = 231 × 293 + 16

⇒ 67699 = 67683 + 16 

⇒ 67699 - 16 = 67683 (completely divisible by 231)

∴ 67683 = 676xy (where x = 8, y = 3)

(3x - 5y) = 3 × 8 - 5 × 3

⇒ 24 - 15 = 9 

∴ The required result = 9

If x2 + ax + b, when divided by x - 5, leaves a remainder of 34 and x2 + bx + a, when divided by x - 5, leaves a remainder of 52, then a + b = ?

  1. 6
  2. -6
  3. 3
  4. -3

Answer (Detailed Solution Below)

Option 1 : 6

Divisibility and Remainder Question 8 Detailed Solution

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x2 + ax + b, when divided by x - 5, leaves a remainder of 34,

⇒ 52 + 5a + b = 34

⇒ 5a + b = 9      ----(1)

Again,

x2 + bx + a, when divided by x - 5, leaves a remainder of 52

⇒ 52 + 5b + a = 52

⇒ 5b + a = 27      ----(2)

From (1) + (2) we get,

⇒ 6a + 6b = 36

⇒ a + b = 6

Find the sum of the numbers between 400 and 500 such that when 8, 12, and 16 divide them, it leaves 5 as remainder in each case.

  1. 922
  2. 932
  3. 942
  4. 912

Answer (Detailed Solution Below)

Option 1 : 922

Divisibility and Remainder Question 9 Detailed Solution

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Calculations:

Numbers are 8, 12 and 16 that must divide numbers between 400 & 500 & get remainder 5

To find the multiple of different numbers, we need to find out the LCM 

LCM of 8, 12, 16

8 = 2³, 12 = 2² × 3, 16 = 2⁴

LCM = 2⁴ × 3 = 48

Number pattern = 48k + 5 (Remainder)

Number between 400 & 500

Smallest number = 48 × 9 + 5 = 437

Largest number = 48 × 10 + 5 = 485

So,

Sum of numbers = 437 + 485

⇒ 922

∴ The correct choice is option 1.

How many numbers are there from 500 to 650 (including both) which are divisible neither by 3 nor by 7?

  1. 87
  2. 99
  3. 121
  4. 21

Answer (Detailed Solution Below)

Option 1 : 87

Divisibility and Remainder Question 10 Detailed Solution

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Given:

The numbers are from 500 to 650 which are divisible neither by 3 nor by 7

Calculation:

The total numbers up to 500 are divisible by 3 = 500/3 → 166 (Quotient)

The total numbers up to 500 are divisible by 7 = 500/7 → 71 (Quotient)

The total numbers up to 500 are divisible by 21 = 500/21 → 23 (Quotient)

The total numbers up to 650 are divisible by 3 = 650/3 → 216 (Quotient)

The total numbers up to 650 are divisible by 7 = 650/7 → 92 (Quotient)

The total numbers up to 650 are divisible by 21 = 650/21 → 30 (Quotient)

⇒ The total number divisible by 3 between 500 and 650 = 216 - 166 = 50

⇒ The total number divisible by 7 between 500 and 650 = 92 - 71 = 21

⇒ The total number divisible by 21 between 500 and 650 = 30 - 23 = 7

The total numbers from 500 to 650 = 150 + 1 = 151

∴ The required numbers = 151 - (50 + 21 - 7) = 151 - 64 = 87

∴ The are total 87 numbers from 500 to 650 (including both) which are neither divisible by 3 nor by 7

What will be the remainder when 2384 is divided by 17?

  1. 1
  2. 2
  3. 3
  4. 4

Answer (Detailed Solution Below)

Option 1 : 1

Divisibility and Remainder Question 11 Detailed Solution

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GIVEN:

2384 is divided by 17.

CALCULATION:

2384 = 2(4 × 96) = 1696

We know that when 16 is divided by 17 the remainder is -1

When 1696 is divided by 17 then remainder = (-1)96 = 1.

A four-digits number abba is divisible by 4 and a

  1. 10
  2. 8
  3. 12
  4. 6

Answer (Detailed Solution Below)

Option 2 : 8

Divisibility and Remainder Question 12 Detailed Solution

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Concept used:

If the last 2 digits of any number divisible by 4, then the number is divisible by 4

Calculation:

According to the question, the numbers are

2332, 2552, 4664, 2772, 6776, 4884, 2992, and 6996

So, there are 8 such numbers in the form abba, divisible by 4

∴ The correct answer is 8

Mistake Points

If you are considering an example ending with 20,

then, 'abba' will be '0220', and 0220 is not a four-digit number. 

Similarly in the case of the example ending with 40,60,80.

If the 5-digit number 750PQ is divisible by 3, 7 and 11, then what is the value of P + 2Q?

  1. 17
  2. 15
  3. 18
  4. 16

Answer (Detailed Solution Below)

Option 1 : 17

Divisibility and Remainder Question 13 Detailed Solution

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Given:

Five-digit number 750PQ is divisible by 3, 7 and 11

Concept used:

Concept of LCM

Calculation:

The LCM of 3, 7, and 11 is 231.

By taking the largest 5-digit number 75099 and dividing it by 231.

If we divide 75099 by 231 we get 325 as the quotient and 24 as the remainder.

Then, the five-digit number is 75099 - 24 = 75075.

The number = 75075 and P = 7, Q = 5

now,

P + 2Q = 7 + 10 = 17

∴ The value of P + 2Q is 17.

If a five digit number 247xy is divisible by 3, 7 and 11, then what is the value of (2y - 8x)?

  1. 9
  2. 17
  3. 6
  4. 11

Answer (Detailed Solution Below)

Option 3 : 6

Divisibility and Remainder Question 14 Detailed Solution

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Given:

If a five digit number 247xy is divisible by 3, 7 and 11

Calculation:

LCM of 3, 7, 11 is 231

According to question

Largest possible value of 247xy is 24799

when we divided 24799 by 231 we get 82 as a remainder

Number = 24799 – 82

⇒ 24717

Now x = 1 and y = 7

(2y – 8x) = (2 × 7 – 8 × 1)

⇒ (14 – 8)

⇒ 6

∴ Required value is 6

What will be the sum of digit of smallest number of four digit which when divided by 16, 19 and 38 leaves the remainder 6 in each case?

  1. 7
  2. 10
  3. 9
  4. 6

Answer (Detailed Solution Below)

Option 1 : 7

Divisibility and Remainder Question 15 Detailed Solution

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Given:

The smallest 4 digit number divided by 16, 19 and 38

and remainder is 6 in each case.

Calculation:

LCM of 16, 19 and 38,

⇒ 16 = 2 x 2 x 2 x 2

⇒ 19 = 19 x 1

⇒ 38 = 2 x 19 x 1

⇒ LCM = 2 x 2 x 2 x 2 x 19 = 304

We know that the smallest number of four digit = 1,000 

When 1,000 divided by 304 then remainder is 88.

So, smallest four digit number which is divided by 304 = 1000 + (304 - 88)

⇒ 1216

Now required number is leaves remainder 6,

so required number = 1216 + 6

⇒ 1222

Sum of digit of 1222 = 1 + 2 + 2 + 2

⇒ 7

∴ Required sum is 7.

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