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

Given:

Total numbers = 90

Numbers are "connected with 9" if:

Divisible by 9

Sum of digits = 9

Contains digit 9

Formula used:

Total numbers not connected with 9 = Total numbers - Numbers connected with 9

Calculation:

Numbers divisible by 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 (10 numbers)

Numbers with sum of digits = 9: (Already counted in divisible by 9)

Numbers containing 9: 19, 29, 39, 49, 59, 69, 79, 89 (8 more numbers)

⇒ Total connected numbers = 10 + 8 = 18

⇒ Numbers not connected with 9 = 90 - 18

⇒ 72

∴ The correct answer is 72.

Given:

The product of two numbers = 9375

The quotient, when the largest number is divided by the smallest number = 15

Formula Used:

Let the numbers be y .

Then, x × y = 9375

And, x/y = 15

Calculation:

Let the two number be x and y where x > y, then

xy = 9375      ...i)

According to the question

x = y × 15

Put the value of x in equation (i)

y × 15 × y = 9375

⇒ y² = 9375/15

⇒ y = 9375/15

⇒ y = √625

⇒ y = 25

From equation (i)

x = 9375/25

⇒ x = 375

∴ Sum of x and y = 375 + 25 = 400

The Correct answer is Option 1

Key Points

The given expression is definitely divisible by 3, as 9 is divisible by 3.

The terms, after division with 3, would be:

9³/3 = 3 × 9²

9⁴/3 = 3 × 9³

9⁵/3 = 3 × 9⁴

……and so on.

All these terms would be odd, as in each of them only odd numbers are getting multiplied.

In total, there would be 98 of these odd terms, whose sum would obviously be even. So, the resultant would be divisible by 2.

So, the original expression is divisible by 3, as well as 2. Which means that it must be divisible by 6 too. So, remainder left would be zero. 

Given:

We need to find the largest two-digit number which, when increased by 5, becomes divisible by both 4 and 5.

Formula Used:

Let the largest two-digit number be x. The condition is x + 5 should be divisible by both 4 and 5.

This implies x + 5 should be divisible by the LCM of 4 and 5, which is 20.

Calculation:

We need x + 5 to be the largest multiple of 20 within the range of two-digit numbers.

The largest two-digit number is 99. So, we start checking from 99 downwards.

For x = 94

⇒ 94 + 5 = 99

99 is not divisible by 20.

For x = 93

⇒ 93 + 5 = 98

98 is not divisible by 20.

For x = 88

⇒ 88 + 5 = 93

93 is not divisible by 20.

For x = 75

⇒ 75 + 5 = 80

80 is divisible by 20.

The largest two-digit number which, when increased by 5, becomes divisible by both 4 and 5 is 75.

The correct answer is option 3.

Given:

Numbers: 1475, 3471, 5418, 4795

Formula Used:

A number is divisible by 9 if the sum of its digits is divisible by 9.

Calculation:

For 1475:

Sum of digits = 1 + 4 + 7 + 5 = 17

17 is not divisible by 9.

For 3471:

Sum of digits = 3 + 4 + 7 + 1 = 15

15 is not divisible by 9.

For 5418:

Sum of digits = 5 + 4 + 1 + 8 = 18

18 is divisible by 9.

For 4795:

Sum of digits = 4 + 7 + 9 + 5 = 25

25 is not divisible by 9.

Correct Option: Option 3

Solution Statement: The number 5418 is divisible by 9 as the sum of its digits (18) is divisible by 9.

Given:

\((49^{15} - 1) \)

Concept used:

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

Here, a & b should be prime number.

Calculation:

\((49^{15} - 1) \)

⇒ \(({(7^2)}^{15} - 1) \)

⇒ \((7^{30} - 1) \)

Here, 30 is a positive integer.

​According to the concept,

\((7^{30} - 1) \) is divisible by (7 + 1) i.e., 8.

∴ 8 is a divisor of \((49^{15} - 1) \).

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

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.