Divisibility and Remainder MCQ Quiz - Objective Question with Answer for Divisibility and Remainder - Download Free PDF
Last updated on Jul 2, 2025
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:
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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) = ?
Answer (Detailed Solution Below)
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.
Answer (Detailed Solution Below)
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.
Answer (Detailed Solution Below)
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 =
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 =
Sum of all terms =
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
Answer (Detailed Solution Below)
Divisibility and Remainder Question 6 Detailed Solution
Download Solution PDFGiven:
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,
∴ 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)?
Answer (Detailed Solution Below)
Divisibility and Remainder Question 7 Detailed Solution
Download Solution PDFGiven:
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 = ?
Answer (Detailed Solution Below)
Divisibility and Remainder Question 8 Detailed Solution
Download Solution PDFx2 + 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.
Answer (Detailed Solution Below)
Divisibility and Remainder Question 9 Detailed Solution
Download Solution PDFCalculations:
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?
Answer (Detailed Solution Below)
Divisibility and Remainder Question 10 Detailed Solution
Download Solution PDFGiven:
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?
Answer (Detailed Solution Below)
Divisibility and Remainder Question 11 Detailed Solution
Download Solution PDFGIVEN:
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
Answer (Detailed Solution Below)
Divisibility and Remainder Question 12 Detailed Solution
Download Solution PDFConcept 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?
Answer (Detailed Solution Below)
Divisibility and Remainder Question 13 Detailed Solution
Download Solution PDFGiven:
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)?
Answer (Detailed Solution Below)
Divisibility and Remainder Question 14 Detailed Solution
Download Solution PDFGiven:
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?
Answer (Detailed Solution Below)
Divisibility and Remainder Question 15 Detailed Solution
Download Solution PDFGiven:
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.