MCQ: Remainder & Divisibility - SSC CGL MCQ

Concept used:

HCF = Highest common factor 

The greatest number which divides each of the two or more numbers is called HCF or Highest Common Factor.

Calculations:

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

H.C.F. of 1651 and 2032

1651 = 13 × 127 

2032 = 4 × 4 × 127

HCF of 1651 and 2032 = 127

∴ The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is 127.

Concept: 

Divisibility rule of  9: A number is divisible by 9, when the sum of the digits of the number is divisible by 9.

Calculation:

5 + m + 8 + 3 + m + 4 + m + 1 

for minimum value 

⇒ 21 + 3m 

⇒ Minimum value of m = 2 (as 27 is the multiple of 9)

For maximum value 

⇒ 21 + 3m      (as 45 is the multiple of 9)

⇒ Maximum value of m = 8

∴ The product of largest and smallest value

⇒ 8 × 2 = 16

Key Points

Here m is a digit (m can be 0, 1, 2, 3,.....,8) not other than this.

We can't take m = 20, m = 30 etc.

Concept used:

If (n + 1)^a is divided by n

Then,

Remainder is 1

Calculation:

7⁴² = (7²)²¹

⇒ (49)²¹

⇒ (48 + 1)²¹

So, the remainder is 1

∴ The required answer is 1.

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 number 5x423y is completely divisible by 88

Concept used:

Divisibility rule of 8 = Last three digit of any number should be divisible by 8 , then number is divisible by 8

Divisibility Rule of 88 = Number should be divisible by both 8 and 11.

Divisibility Rule of 11 = Sum of odd place digit - Sum of even digit place = 0 or multiple of 11

Calculation:

If a number 5x423y is completely divisible by 88

So, By using divisibility rule of 8

y = 2 because 232 is divisible by 8

Now, By using divisibility rule of 11

⇒ (x + 2 + 2) - (5+4+3) = 0

⇒ (x + 4) - (12) = 0

⇒ x = 8

The value of 5x - 8y = 40 - 16 = 24

∴ The required answer is 24

Numbers that are divisible by 7 between 244 and 332,

⇒ 245, 252.........329

Using AP:

a_n = a + (n - 1)d

329 = 245 + (n - 1) × 7

n = 13

∴ There are 13 numbers

Given:

2⁸⁹ is divided by 9.

Concept:

If we divide 8 by 9 we will get the remainder as - 1

But remainder can not be negative 

Therefore we can not use -1 as the remainder

To simplify calculation, we use 9 – 1 = 8 as a remainder

Calculation:

(2⁸⁹/9)

⇒ {(2³)²⁹ × 2²}/9

⇒ {(8)²⁹ × 4}/9  

⇒ {(-1)²⁹ × 4}/9

⇒ (-1 × 4)/9

⇒ - 4/9

⇒ Remainder = - 4 + 9 = 5

∴ The remainder if 289 is divided by 9 is 5.

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.

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:

(49¹⁵ − 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:

(49¹⁵ − 1)

⇒ ((72)¹⁵ − 1)

⇒ (730 − 1)

Here, 30 is a positive integer.

​According to the concept,

(7³⁰ − 1) is divisible by (7 + 1) i.e., 8.

∴ 8 is a divisor of (49¹⁵ − 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

Given:

The divisor is 22.8 and the quotient is 8.5.

Remainder = 0

Concept used:

Dividend = Divisor × Quotient + Remainder

Calculation:

According to the question, the dividend

⇒ 22.8 × 8.5 + 0

⇒ 193.8

∴ The dividend is 193.8.

Given:

The 9-digit number 83x93678y is divisible by 72.

Concept used:

If the last three digits of a number are divisible by 8, then the number is completely divisible by 8.

If the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.

Calculation:

If the 9-digit number 83x93678y is divisible by 72, then it must be divisible by 8 and 9 simultaneously.

The last three digits of 83x93678y are 78y.

Hence, In order to be divisible by 8, y must be 4.

The number becomes = 83x936784

Now, in order to be divisible by 9,

8 + 3 + x + 9 + 3 + 6 + 7 + 8 + 4 = Multiple of 9

⇒ 48 + x = Multiple of 9

In order to be a multiple of 9, x must be 6. That makes (48 + 6) i.e. 54 a multiple of 9.

Hence,

⇒ 3x - 2y

⇒ 3 × 6 - 2 × 4

⇒ 10

∴ The value of (3x - 2y) is 10.

Given:

The eight-digit number is 6181a12b.

Concept used:

Divisibility rule for 8: The last three-digit of any number is divisible by 8.

Divisibility rule for 9: The sum of digits of the given number is divisible by 9.

Divisibility rule for 72: The required number is divisible by 8 and 9 then the number is divisible by 72.

Calculation:

As we know, If the required number 6181a12b is divisible by 8 and 9 then the number is divisible by 72.

As per the divisibility rule for 8, the last three-digit of any number is divisible by 8. i.e., 12b must be divided by 8.

The possible value of b: 0, 8

As per the divisibility rule for 9,

For b = 0:

⇒ (6 + 1 + 8 + 1 + a + 1 + 2 + 0) divisible by 9.

⇒ (19 + a) divisible by 9.

Therefore, the required value of 'a' is 8 so that the required number is divisible by 9.

For b = 8:

⇒ (6 + 1 + 8 + 1 + a + 1 + 2 + 8) divisible by 9.

⇒ (27 + a) divisible by 9.

Therefore, the required value of 'a' can be 0 and 9 so that the required number is divisible by 9.

The possible value of (a + b)² can be:

⇒ (a + b)² = (8 + 0)² = 8² = 64

⇒ (a + b)² = (0 + 8)² = 8² = 64

⇒ (a + b)² = (8 + 9)² = 17² = 289

The greatest value of (a + b)² is 289.

∴ The greatest value of (a + b)² is 289.

Given:

The eight-digit number is 1862383A.

Concept used:

Divisibility rule for 2: The last digit of any digit number is an even number.

Divisibility rule for 11: The difference for the sum of the digits in the odd and even places of the given number is 0 or divisible by 11.

Divisibility rule for 22: The required number is divisible by 2 and 11 then the number is divisible by 22.

Calculation:

Prime factorization of 22 = 2 × 11

As we know, If the required number 1862383A is divisible by 2 and 11 then the number is divisible by 22.

As per the divisibility rule for 2, the value of A must be an even number.

The possible value of A: 2, 4, 6, 8, 0

As per the divisibility rule for 11,

⇒ (8 + 2 + 8 + A) - (1 + 6 + 3 + 3) = 0 or multiple of 11

⇒ 18 + A - 13 = 0 or multiple of 11

⇒ 5 + A = 0 or multiple of 11

For 5 + A to be a multiple of 11, the value of A must be 6.

∴ The required value of A is 6.