Divisibility Questions and Answers

March 6, 2024

Divisibility Questions and Answers in Aptitude

On this page we have discussed about the Divisibility Question and answer along with the  definition and rules and question with detailed explanation. 

Divisibility Questions and Answers

Rules of Divisibility  : 

Divisibility rule of 2Any number whose last digit is an even number (0, 2, 4, 6, 8) is divisible by 2
Divisibility rule of 3A number is divisible by 3 if the sum of its digits is divisible by 3.
Divisibility rule of 4A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
Divisibility rule of 5A number is exactly divisible by 5 if it has the digits 0 or 5 at one’s place.
Divisibility rule of 6A number is exactly divisible by 6 if that number is divisible by 2 and 3 both. It is because 2 and 3 are prime factors of 6.
Divisibility rule of 7Double the last digit and subtract it from the remaining leading truncated number to check if the result is divisible by 7 until no further division is possible
Divisibility rule of 8If the last three digits of a number are divisible by 8, then the number is completely divisible by 8.
Divisibility rule of 9It is the same as of divisibility of 3. Sum of digits in the given number must be divisible by 9.
Divisibility rule of 11If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11.
Divisibility rule of 12A number is exactly divisible by 12 if that number is divisible by 3 and 4 both.
Divisibility rule of 13Multiply the last digit with 4 and add it to remaining number in a given number, the result must be divisible by 13.
Divisibility rule of 15If the number divisible by both 3 and 5, it is divisible by 15.
Divisibility rule of 17Multiply the last digit with 5 and subtract it from remaining number in a given number, the result must be divisible by 17
Divisibility rule of 19Multiply the last digit with 2 and add it to remaining number in a given number, the result must be divisible by 19.

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Divisibility Questions and Answers

1. Calculate the remainder for the number 3697 when it is divided by 2?

2

2

3.41%

5

5

3.12%

1

1

92.48%

10

10

0.99%

When we divide 3697 by 2, we get 1848 as the quotient and 1 as the remainder.

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2. What is the quotient of the given number 2368 when it is divided by 3? 

7136

7136

3.88%

368

368

2.59%

789

789

91.45%

956

956

2.08%

So 789 will be Quotient

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3. Which number among the given option is not divisible by 3?

111

111

7.96%

993

993

1.52%

125

125

89.28%

999

999

1.25%

The simple way is to sum all the digits, if the result is divisible by 3, the number is divisible by 3.

1 + 1 + 1 = 3

9 + 9 + 3 = 21

1 + 2 +5 = 8

9 + 9 + 9 = 27

All the numbers above are divisible by 3, except 8.

Therefore, 125 is not divisible by 3.

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4. Calculate the smallest integer which is a multiple of 3, 5, and 10.

52

52

2.44%

20

20

4.85%

30

30

91.43%

35

35

1.28%

The LCM of 3, 5, and 10 is 30.

Therefore option C is the correct one.

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5. Calculate how many numbers less than 500 are divisible by 2 or 3 or 4 or 5 or 7 or 21?

457

457

17.56%

368

368

15.71%

250

250

25.45%

336

336

41.28%

Step 1: Find the counts of numbers divisible by each individual divisor.

Divisible by 2: There are 500 / 2 = 250 numbers divisible by 2.
Divisible by 3: There are 500 / 3 = 166 numbers divisible by 3.
Divisible by 4: There are 500 / 4 = 125 numbers divisible by 4.
Divisible by 5: There are 500 / 5 = 100 numbers divisible by 5.
Divisible by 7: There are 500 / 7 = 71 numbers divisible by 7.
Divisible by 21: There are 500 / 21 = 23 numbers divisible by 21.
Step 2: Apply inclusion-exclusion principle.

We need to find the total count of numbers divisible by at least one of the given divisors. We add the counts from step 1 and then subtract the counts of overlaps (divisible by multiple divisors).

Total count = (Divisible by 2) + (Divisible by 3) + (Divisible by 4) + (Divisible by 5) + (Divisible by 7) + (Divisible by 21)
Total count = 250 + 166 + 125 + 100 + 71 + 23 = 735

However, we've counted some numbers multiple times due to their divisibility by multiple divisors. Let's correct for overlaps:

Divisible by 2 and 3: There are 500 / (2 * 3) = 83 numbers divisible by both 2 and 3.

Divisible by 2 and 4: There are 500 / (2 * 4) = 62 numbers divisible by both 2 and 4.

Divisible by 2 and 5: There are 500 / (2 * 5) = 50 numbers divisible by both 2 and 5.

Divisible by 2 and 7: There are 500 / (2 * 7) = 35 numbers divisible by both 2 and 7.

Divisible by 2 and 21: There are 500 / (2 * 21) = 11 numbers divisible by both 2 and 21.

Divisible by 3 and 4: There are 500 / (3 * 4) = 41 numbers divisible by both 3 and 4.

Divisible by 3 and 5: There are 500 / (3 * 5) = 33 numbers divisible by both 3 and 5.

Divisible by 3 and 7: There are 500 / (3 * 7) = 23 numbers divisible by both 3 and 7.

Divisible by 3 and 21: There are 500 / (3 * 21) = 7 numbers divisible by both 3 and 21.

Divisible by 4 and 5: There are 500 / (4 * 5) = 25 numbers divisible by both 4 and 5.

Divisible by 4 and 7: There are 500 / (4 * 7) = 17 numbers divisible by both 4 and 7.

Divisible by 4 and 21: There are 500 / (4 * 21) = 5 numbers divisible by both 4 and 21.

Divisible by 5 and 7: There are 500 / (5 * 7) = 14 numbers divisible by both 5 and 7.

Divisible by 5 and 21: There are 500 / (5 * 21) = 4 numbers divisible by both 5 and 21.

Divisible by 2, 3, and 4: There are 500 / (2 * 3 * 4) = 41 numbers divisible by 2, 3, and 4.

Divisible by 2, 3, and 5: There are 500 / (2 * 3 * 5) = 27 numbers divisible by 2, 3, and 5.

Divisible by 2, 3, and 7: There are 500 / (2 * 3 * 7) = 11 numbers divisible by 2, 3, and 7.

Divisible by 2, 3, and 21: There are 500 / (2 * 3 * 21) = 3 numbers divisible by 2, 3, and 21.

Divisible by 2, 4, and 5: There are 500 / (2 * 4 * 5) = 25 numbers divisible by 2, 4, and 5.

Divisible by 2, 4, and 7: There are 500 / (2 * 4 * 7) = 8 numbers divisible by 2, 4, and 7.

Divisible by 2, 4, and 21: There are 500 / (2 * 4 * 21) = 2 numbers divisible by 2, 4, and 21.

Divisible by 3, 4, and 5: There are 500 / (3 * 4 * 5) = 8 numbers divisible by 3, 4, and 5.

Divisible by 3, 4, and 7: There are 500 / (3 * 4 * 7) = 4 numbers divisible by 3, 4, and 7.

Divisible by 3, 4, and 21: There are 500 / (3 * 4 * 21) = 1 number divisible by 3, 4, and 21.

Divisible by 4, 5, and 7: There are 500 / (4 * 5 * 7) = 3 numbers divisible by 4, 5, and 7.

Divisible by 4, 5, and 21: There are 500 / (4 * 5 * 21) = 1 number divisible by 4, 5, and 21.

Divisible by 2, 3, 4, and 5: There are 500 / (2 * 3 * 4 * 5) = 4 numbers divisible by 2, 3, 4, and 5.

Divisible by 2, 3, 4, and 7: There are 500 / (2 * 3 * 4 * 7) = 2 numbers divisible by 2, 3, 4, and 7.

Divisible by 2, 3, 4, and 21: There are 500 / (2 * 3 * 4 * 21) = 1 number divisible by 2, 3, 4, and 21.

Divisible by 2, 4, 5, and 7: There are 500 / (2 * 4 * 5 * 7) = 1 number divisible by 2, 4, 5, and 7.

Divisible by 2, 4, 5, and 21: There are 500 / (2 * 4 * 5 * 21) = 1 number divisible by 2, 4, 5, and 21.

Divisible by 3, 4, 5, and 7: There are 500 / (3 * 4 * 5 * 7) = 1 number divisible by 3, 4, 5, and 7.

Divisible by 3, 4, 5, and 21: There are 500 / (3 * 4 * 5 * 21) = 0 numbers divisible by 3, 4, 5, and 21.

Divisible by 2, 3, 4, 5, and 7: There are 500 / (2 * 3 * 4 * 5 * 7) = 0 numbers divisible by 2, 3, 4, 5, and 7.

Now we can subtract the counts of overlaps from the total count:

Total count = 735 - (83 + 62 + 50 + 35 + 11 + 41 + 33 + 23 + 7 + 25 + 17 + 5 + 14 + 4 + 41 + 27 + 11 + 3 + 8 + 4 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1)

Total count = 735 - 399 = 336

So, there are 336 numbers less than 500 that are divisible by 2, 3, 4, 5, 7, or 21.

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6. Identify the odd number among the following.

21

21

83.08%

44

44

9.7%

644

644

3.72%

70

70

3.5%

When we divide all the given options by 2, we get remainder only in the case of 21.

All the remaining numbers generate 0 as the remainder. Thus the odd number in the given options is 21.

Therefore Option A is the correct one.

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7. Calculate the quotient and the remainder when 3472 is divided by 7.

496, 0

496, 0

90.3%

500, 5

500, 5

1.62%

456, 2

456, 2

5.42%

789, 1

789, 1

2.66%

The quotient is 496, and the remainder is 0.

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8. Calculate the probability of divisibility of the sum of the first B number by B.

\frac{1}{2}

\frac{1}{2}

78.71%

\frac{3}{4}

\frac{3}{4}

10.55%

\frac{5}{7}

\frac{5}{7}

4.61%

\frac{1}{9}

\frac{1}{9}

6.13%

We already know that the sum of the first n numbers is \frac{B (B+1)}{2}

If B is an even number, the probability will be zero.

If B is odd, the probability is 1

If we consider the first B numbers, then the probability will be \frac{1}{2}

Therefore Option A is the correct one.

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

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9. Calculate how many numbers between 1 and 100, including both are divisible by 9 or 4.

34

34

76.28%

23

23

9.84%

45

45

8.29%

13

13

5.58%

11 numbers are divisible between 1 and 100 by 9.

25 numbers are divisible by 4 between 1 and 100.

Therefore, total numbers = 36.

Also, there are numbers which are divisible by both 4 and 9 and are counted twice.

= 11 + 25 – 2 = 34.

Therefore option A is the correct one.

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

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10. Calculate the smallest five digit number divisible by 12, 13, and 20.

10247

10247

6.61%

10140

10140

75.68%

12361

12361

4.83%

None of the above

None of the above

12.88%

LCM of 12, 13, and 20 is 780.

The least five digit number is 10000

10000/780 = 12.82

Therefore, the 13th multiple must be at least 5 digit number divisible by 12, 13, and 20.

= 780 x 13 = 10140

Hence Option B is the correct one.

Good Job! Did you know? : With $8.5 billion in their kitty, Wipro is the 4th largest Indian IT company in terms of revenue.

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