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

Given:

9-digit number: 5x79856y6

This number is divisible by 36.

x and y are natural numbers.

Formula Used:

A number is divisible by 36 if it is divisible by both 4 and 9.

1. For divisibility by 4, the last two digits must form a number divisible by 4.

2. For divisibility by 9, the sum of all digits must be divisible by 9.

Calculation:

Last two digits are 6y6, so for divisibility by 4:

y6 must be divisible by 4.

Possible values for y: 1, 3, 5, 7, 9 (natural numbers)

Sum of digits: 5 + x + 7 + 9 + 8 + 5 + 6 + y + 6 = 46 + x + y

For divisibility by 9, (46 + x + y) must be divisible by 9.

Considering y = 9 (largest possible value):

46 + x + 9 = 55 + x

55 + x must be divisible by 9.

Possible values for x = 8 (as 55 + 8 = 63)

so, x = 8 and y = 9

Now,

Neagtive value of  √(2x + y) =  √(2 × 8 + 9) =  √25 = - 5

The correct answer is option 4.

Given:

If n is a natural number, then 92ⁿ - 42ⁿ is always divisible by:

Options:

1) 5

2) 13

3) Both (A) and (B)

4) Neither (A) nor (B)

Formula used:

For divisibility rules, we check if the expression satisfies modulo conditions for the given numbers.

92 ≡ -3 (mod 5), 42 ≡ 2 (mod 5)

92 ≡ 1 (mod 13), 42 ≡ 3 (mod 13)

Calculations:

Step 1: Modulo 5 check

92ⁿ ≡ (-3)ⁿ (mod 5), 42ⁿ ≡ 2ⁿ (mod 5)

⇒ 92ⁿ - 42ⁿ ≡ (-3)ⁿ - 2ⁿ (mod 5)

For all n, (-3)ⁿ ≡ 2ⁿ (mod 5), hence divisible by 5.

Step 2: Modulo 13 check

92ⁿ ≡ 1ⁿ (mod 13), 42ⁿ ≡ 3ⁿ (mod 13)

⇒ 92ⁿ - 42ⁿ ≡ 1ⁿ - 3ⁿ (mod 13)

For all n, 1ⁿ - 3ⁿ ≡ 0 (mod 13), hence divisible by 13.

Step 3: Combine results

Since 92ⁿ - 42ⁿ is divisible by both 5 and 13, it is divisible by both.

∴ The correct answer is option (3).

Given:

Dividend = 2200

Remainder = 13

Divisor = one-third of the Quotient

Formula used:

Dividend = Divisor × Quotient + Remainder

Divisor = Quotient ÷ 3

Calculation:

2200 = Divisor × Quotient + 13

⇒ 2200 - 13 = Divisor × Quotient

⇒ 2187 = Divisor × Quotient

Divisor = Quotient ÷ 3

⇒ 2187 = (Quotient ÷ 3) × Quotient

⇒ 2187 = Quotient² ÷ 3

⇒ Quotient² = 2187 × 3

⇒ Quotient² = 6561

⇒ Quotient = √6561

⇒ Quotient = 81

Divisor = Quotient ÷ 3

⇒ Divisor = 81 ÷ 3

⇒ Divisor = 27

∴ The correct answer is option (2).

Given:

The sum of the digits of a four-digit number 'abcd' is subtracted from the number.

We need to find the number by which the result is always divisible.

Formula used:

Let the number be represented as: (1000a + 100b + 10c + d)

Sum of digits = (a + b + c + d)

Result after subtraction = (1000a + 100b + 10c + d) - (a + b + c + d)

Calculation:

Result = (1000a + 100b + 10c + d) - (a + b + c + d)

⇒ (1000a - a) + (100b - b) + (10c - c) + (d - d)

⇒ (999a + 99b + 9c)

Take common factors:

⇒ (9 × (111a + 11b + c))

This shows that the result is always divisible by 9.

∴ The correct answer is option (1).

Given:

85 × 87 × 89 × 91 × 95 × 96

Calculations:

85 × 87 × 89 × 91 × 95 × 96

⇒ (17 × 5) × 87 × 89 × (19 × 5) × (24 × 4)

⇒ 17 × 87 × 89 × 19 × 24 × (5 × 5 × 4)

⇒ 17 × 87 × 89 × 19 × 24 × 100

The expression is divided by 100,

⇒ 17 × 87 × 89 × 19 × 24 × 100/100 = 17 × 87 × 89 × 19 × 24

Since the expression is completely divisible by 100: Remainder = 0

∴ The remainder when 85 × 87 × 89 × 91 × 95 × 96 is divided by 100 is 0.

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 .

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

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

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.

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

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.

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.

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.

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

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.