### 01 Jun Number system has a special place in quantitative aptitude section of Banking and SSC exams. Within this area, there are a lot of topics such as remainder theorem, cyclicity of numbers, divisibility rules etc. Each of these topics is important in its own right and must not be neglected. When it comes to divisibility rules, the broad scope of application makes them highly critical from exam perspective.

Therefore, experts from the institute for Best IBPS PO Coaching in Delhi have enlisted the divisibility rules that you must learn before appearing in any competitive exam. These rules and their application are as follows:

2: If the last digit is even, the number is divisible by 2.

3: If the sum of digits is divisible by 3, then so is the number.

4: If the last 2 digits of a number are divisible by 4, then so is the number.

5: If the last digit is a 5 or a 0, the number is divisible by 5.

6: If the number is divisible by both 3 & 2, then it is divisible by 6 as well.

8: If the last 3 digits of a number are divisible by 8, then so is the number.

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

10: If the number ends in 0, it is divisible by 10.

11: If you add every second digit & then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11.

12: If the number is divisible by both 3 and 4, it is also divisible by 12.

25: Numbers that end in 00, 25, 50, or 75 are divisible by 25.

Question: 1. What is the remainder when 8078×16085×16868 is divided by 8? (a) 1 (b) 3 (c) 4 (d) 0

Solution: (d)

The divisibility rule of 8 is to divide the last 3 digits of any number by 8. If the last 3 digits are divisible by 8, then the number is completely divisibly by 8, otherwise not. So, if the number is not divisible by 8, we can easily get remainders.

But, over here are three numbers in multiplication. We cannot multiply such big numbers, so it seems very difficult to find the remainder. Here, we can directly divide each of these three numbers by 8 & then multiply their remainders to get the final remainders.

• Dividing 8078 by 8, we will get 6 as the remainder
• Dividing 16085 by 8, we will get 5 as the remainder
• Dividing 16868 by 8, we will get 4 as the remainder

Multiplying all three remainders 6×5×4, we get 120. But, this cannot be the remainder as 120 is greater than 8. So, we will further divide 120 by 8 and finally get 0 as the remainder.

Question: 2. What will be the remainder if (1! + 2! + 3! + ….+ 100!) is divided by 8? (a) 1 (b) 2 (c) 3 (d) 4

Solution: (a)

1! = 1, 2! = 2×1 = 2, 3! = 3×2×1 = 6, 4! = 4×3×2×1 = 24, 5! = 5×4×3×2×1 = 120 & so on. Here, we can see that after 3!, every number is divisible by 8 like 4!, 5!, 6!…… Only the first three numbers (i.e. 1!, 2! & 3!) are not divisible by 8. This obviously means that they will give us remainders.

Now, if (1! + 2!  +3! + …….+ 100!) is divided by 8 or (1+2+6+24+120+………..) is divided by 8, the remainder will come only from the first three numbers = 1 + 2+ 6 = 9. Further dividing 9 by 8, the final remainder will be 1.

Question: 3. Which of the following numbers is divisible by 99? (a) 135732 (b) 803358 (c) 3572403 (d) 114345

Solution: (d)

A number is divisible by 99, only when it is divisible by both 11 & 9. Upon checking each option for divisibility with both 11 & 9, we find that only the 4th option 114345 is such that it is divisible by both 9 & 11. 1+1+4+3+4+5 = 18, so the number is divisible by 9. 1+4+4 – (1+3+5) = 9 – 9 = 0.

Question: 4. What least number must be added to 5647132 to make it divisible by 9? (a) 1 (b) 2 (c) 3 (d) None of these

Solution: (d)

The rule of 9 is that the sum of all digits must be divisible by 9. Here, the sum of all digits of 5647132 = 28. Now, 28 is not divisible by 9. Therefore, 8 must be added to 28. This way it will become 36, which is divisible by 9. Hence, the answer is 8.

Question: 5. Find the value of * in the number 4*56 so as to make it divisible by 33 (a) 3 (b) 4 (c) 5 (d) None of these

Solution: (a)

4*56 is divisible by 33. So, it must be divisible by both 11 & 3. By applying the rule of 11, 4+5 – (*+6) = 0 or 9 – (*+6) = 0 or 3 – * = 0 or * = 3.

Question: 6. A number 1568a35b is divisible by 88. Find a & b. (a) 3,2 (b) 4,6 (c) 6,2 (d) 2,8

Solution: (c)

For the number 1568a35b to be divisible by 88, it will have to be divisible by both 11 & 8. Applying the rule of 8, we can say that this number’s last 3 digits are divisible by 8 or 35b is divisible by 8. On dividing 35b by 8, it is clear that b = 2 as we know that there is only one number 32 in line of 30 which is divisible by 8. Now, applying the rule of 11 in given number 1568a352, sum of odd places – sum of even numbers = 0

or 1+6+a+5 –(5+8+3+2)= 0

→ 12+a – 18 = 0

→ a – 6 = 0

→ a = 6.

Hence, the values of a & b are 6 & 2 respectively.

Rule 1: For finding the greatest number divisible by a certain number, subtract the remainder from the given greatest number.

OR

The least number that must be subtracted from any number to make it exactly divisible by a certain divisor is = remainder.

Rule 2: For finding the least number divisible by a certain number, add (divisor – remainder) to the given least number.

OR

The least number that must be added to any number to make it exactly divisible by a certain divisor is = (divisor – remainder).

Rule 3: For finding the nearest number to any given number divisible by a certain number, check whether subtraction of remainder is close to the given number or addition of difference between divisor & remainder is close to the given number.

Question: 7. What is the least number that must be subtracted from 7963 to make it exactly divisible by 65? (a) 31 (b) 32 (c) 33 (d) 34

Solution: (c)

On dividing 7963 by 65, we get 33 as remainder. So, the number to be subtracted will be 33.

Question: 8. What least number should be added to 7963 so that it is exactly divisible by 65? (a) 12 (b) 32 (c) 42 (d) 50

Solution: (b)

When 7963 is divided by 65, the remainder is 33, So, the number that must be added is 65 33 = 32

Question: 9. Find the smallest number of five digits which is exactly divisible by 73. (a) 10001 (b) 10002 (c) 10003 (d) 10004

Solution: (a)

The smallest five digit number is 10000. On dividing 10000 by 73, we get 72 as remainder. So, we will add divisor – remainder = (73 – 72) = 1

Hence, the least number of 5 digits which is exactly divisible by 73 will be = 10000 + 1 = 10001

Question: 10. Find the greatest number of five digits which is exactly divisible by 147. (a) 99960 (b) 99963 (c) 99968 (d) 99960

Solution: (d)

The greatest number of five digits is 99999. On dividing it by 147, the remainder is 39. So, the required number will be 99999 39 = 99960.

Question: 11. Find the number nearest to 3105 that is exactly divisible by 21. (a) 3087 (b) 3102   (c) 3108    (d) None of these

Solution: (c)

On dividing 3105 by 21, we get 18 as remainder. So, the number to be subtracted so that 3105 becomes divisible by 21 is = 18. However, the number to be added so that 3105 becomes divisible by 21 is = 21 – 18 = 3. So, addition of 3 is closer as compared to subtraction of 18. Hence, nearest number to 3105 that is exactly divisible by 21 = 3105 + 3 = 3108

Question: 12. Find the nearest number to 6444 that is exactly divisible by 42. (a) 6422 (b) 6426 (c) 6420 (d) 6428

Solution: (b)

On dividing 6444 by 42, we get 18 as remainder. So, the number to be subtracted so that 6444 becomes divisible by 42 is = 18 & the number to be added so that 6444 will be exactly divisible by 42 = 42 – 18 = 24.

We know that subtraction of 18 is closer as compared to addition of 24. Hence, the number nearest to 6444 that is exactly divisible by 42 is = 6444 – 18 = 6426.

Finally, it is suggested that to improve your understanding of number system, you must attempt a lot of practice questions. In this regard, the study material provided by the Top SSC Coaching Centre in Delhi can be really beneficial to you. Once you are through with the practice material, do make it a point to attempt sufficient number of mock tests. Doing so will surely improve your speed and accuracy.

Summary

This article on number system covers fundamentals and application of all the divisibility rules that are important from the point of view of SBI PO and IBPS Clerk exam preparation. If you have any doubts and queries, pls. feel free to get in touch with Vidya Guru Experts by writing at vidyagurudelhi@gmail.com.