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Divisibility concept is one of the most important and frequently tested topics in quantitative aptitude and number theory for competitive exams. It is asked in almost every major exam including SSC CGL, SSC CHSL, Bank PO, Bank Clerk, Railway RRB, CAT and UPSC CSAT. This topic is also important for GMAT, GRE, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests worldwide. A strong understanding of divisibility rules, divisibility tests for all numbers from 2 to 17 and successive division is essential for scoring well in these exams. In this post we cover everything from classification of numbers, divisibility rules for 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13 and 17 with multiple methods, general properties of divisibility and successive division concept — all explained with solved examples.

📚 What You Will Learn in This Post

Classification of Numbers — Natural, Whole, Integer, Rational, Irrational and Complex

Divisibility Rules for 2, 3, 4, 5, 6, 8, 9, 10 and 11

Divisibility by 7 — Three Methods with Solved Examples

Divisibility by 13 and 17 — Methods and Solved Examples

General Properties of Divisibility — Important Rules

Successive Division — Concept and Solved Examples

Solved Examples on Divisibility for Competitive Exams

                                                 Divisibility

Number & their classification:-

Natural Number(N):-   {1, 2, 3, 4, 5, …………….}
Whole Number(W):-  {0, 1, 2, 3, 4, 5, ……………. }
Integer(Z):-  {……. -4, -3, -2, -1, 0, 1, 2, 3, ……….}
Rational Number(Q):- \(\frac{1}{2}\), \(\frac{3}{7}\), \(\frac{1}{2}\), \(\frac{-6}{5}\), 0, +5, -130, ………..
Irrational Number:-  π, \(\sqrt 2 \), \(\sqrt 5 \), \(\sqrt 7 \), …………
Complex Number:-\(\sqrt { – 8} \), \(\sqrt { – 7} \), \(\sqrt { – 5} \), \(\sqrt { – 1} \), \(2 + \sqrt { – 5} \), \(3 + \sqrt { – 7} \), …………
Test for divisibility of numbers

Divisibility by 2:- If the last digit is an even number or it has 0 at the end then the number is divisible by 2.
Ex: 20, 50, 512, 678, 980324, 278

Divisibility by 3:- If the sum of the digits of the given number is divisible by 3 then the number is divisible by 3

Ex:- 513, 900, 87021, 2345Sum of digits991821

Divisibility by 4:- If the number formed by last two digits of a number is divisible by 4 then the number is divisible by 4.

Ex: -432, 500, 816, 512, 780030572

Divisibility by 5:- If the last digit of the given number is 0 or 5 then the number is divisible by 5.

Ex: -60, 5780, 6635, 78925, 7080900 or 5

Divisibility by 6:- If the given number is divisible by both 2 and 3 then the number is divisible by 6.

Ex: -8760, 23052 , 90070824ZeroEvenEvenSum of digits 21 12 30

Divisibility by 7:-

Method (1):
Take the last digit of the number an multiply it by 2.
Subtract the result from the rest of the number
Repeat the process if necessary until you get a small number.
If the result is 0 or multiple of 7, the original number is divisible by 7 otherwise, it is not.

Ex:- 6 1 9 1 5 8 4×286 1 9 1 5 0×2- 06 1 9 1 5×2- 106 1 8 1×2- 26 1 6×2- 1249Divisible by 7Hence given numberis divisible by 7 Answer

Method(2): Adding 5 times the last digit to the rest gives a multiple of 7. 

Ex:- 9 6 1 4 3 5 2 0 9×2-189 6 1 4 3 5 0 2×2- 49 6 1 4 3 4 6×2- 129 6 1 4 2 2×29 6 1 3 8- 4×2- 169 5 9 7×2- 149 4 1 5Divisible by 7Hence given numberis divisible by 7 Answer×2- 1084

Method(2): Adding 5 times the last digit to the rest gives a multiple of 7.

Ex:- 6 1 9 1 5 8 2×5+106 1 9 1 6 8×5+ 406 1 9 5 6×5+ 306 2 2 5×5+106 3 2×5+ 1073Not divisible by 7Hence given numberis divisible by 7 AnswerEx:- 9 6 1 4 3 5 2 0 9×5+109 6 1 4 3 5 6 5×5+ 259 6 1 4 3 8 1×5+ 5×5+159 6 6 0×509 6 6Divisible by 7Hence given numberis divisible by 7 Answer9 6 1 4 4 3×5+ 30126Ex:- 6 1 9 1 5 8 4×5+206 1 9 1 7 8×5+ 406 1 9 5 7×5+ 356 2 3 0×5+06 2 3×5+ 1577Divisible by 7Hence given numberis divisible by 7 Answer

Method(3):
Using three-digit pairs:-
⟶ Split the number into groups of three digits from right to left.
⟶ calculate the alternating sum of these groups
⟶ If the result is 0 or divisible by 7, the original number is divisible by 7.    

Ex: 6191584 961435209 9614352090+ 6 - 191 + 584 = 399+ 961 - 435 + 209 = 735Divisible by 7Divisible by 7Hence original numberis divisible by 7Hence original numberis divisible by 7++--= 343Divisible by 7Hence original numberis divisible by 7

Divisibility by 8:- If the number formed by the last three digits of the given number is divisible by 8 then the given number is divisible by 8.

Ex: 7512 67008 382616divisible by 8divisible by 8divisible by 8

Divisibility by 9: If the sum of digits of the given number is divisible by 9.

Ex: 23085 96327 1233Sum of digits18279

Divisibility by 10:- If the last digit of the number is 0.
Ex:   680         57320            4100850

Divisibility by 11:- If the difference of the sum of its digits in odd places and the sum of its digits in even places is either 0 or multiplee by 11.

Ex: 3 6 2 3 415432+9+9-0Number is divisible by 116 2 9 1 3 1 8264++-22Hence number is divisible by 11divisible by 11

Divisibility by 13:-
Method (1):
 Add 4 times the last digit to the rest. The result must be 0 or divisible by 13.

Ex:- 7 5 9 2×4+87 6 7×4+ 28104Divisible by 13Ex:- 3 4 5 4 4×4+163 4 7 0×403 4 7Divisible by 13×42852

Method(2): Subtract 9 times the last ddigit from the rest. The result must be 0 or divisible by 13.

Ex:- 6 4 8 3 7 5×9- 456 4 7 9 2×9- 186 4 6 1×9- 96 3 7×9- 630ZeroHence number is divisible by 13Ex:- 1 2 3 3 8 3 5 2×9- 181 2 3 3 8 1 7×9- 631 2 3 3 1 8×9- 721 2 2 5 9×9- 81Divisible by 131 1 4 4×93678Hence number is divisible by 13

Method(3):
⟶ Split the number into groups of three from right to left
⟶ calculate the alternating sum of these groups
⟶ If the result is 0 or divisible by 13, the original number is divisible by 13

Ex: 7660068585++--= - 7 + 660 - 068 + 585= 1170Divisible by 13∴ original number is divisible by 13

Divisibility by 17:-
Method(1):
 Subtract 5 times the last digit from the rest. The result must be 0 or divisible by 17.

Ex:- 1 8 2 2 1 4 5×5- 251 8 2 1 8 9×5- 451 8 1 7 3×5- 151802Divisible by 17Hence original number isdivisible by 17

Method(2): Add 12 times the last digit to the rest. The result must be 0 or divisible by 17.

Ex:- 1 8 2 2 1 4 5×12+ 601 8 2 2 7 4×12+ 481 8 2 7 5×12601 8 8 7Divisible by 17Hence original number isdivisible by 17×1284272

Some General Property of Divisibility:-

● If a number x is divisible by y , then any number divisible by x, will also be divisible by y and all the factors of y.
Ex: The number 72 is divisible by 6. Thus any number that is divisible by 72, will also be divisible by 6 an also factors of 6 i.e. by 2 and 3.
● If a number x is divisible by two or more than two co-prime numbers then x is also divisible by the product  of those numbers.
Ex:- The number 2040 is divisible by 3, 8, 17 that are co-prime to each other so 2040 will also be divisible by 3 × 8 = 24, 3 × 17 = 51, 8 × 17 = 136
● If two numbers x and y are divisible by ‘p’, then their sum x + y is also divisible by p.
Ex:- The number 138 and 312 are both divisible by 6. Thus their sum = 138 + 312  = 450 will also be divisible by 6.
● If any number is written 6 times like 111111, 222222, 777777 etc, it will be exactly divisible by 7, 11, 13, 37 & 1001
● Any number written like 2727, 3535, 2929 will be divisible by 101.

(a - b)n is even number this number is divisible by (a - b) & (a + b)n is odd number this number is divisible by (a + b)

(aⁿ + bⁿ) n is odd then this number is divisible by (a + b)
(aⁿ + bⁿ) n is even then it can not be determined.
● A six digit number is formed by repeating a three digit number; for example 257257, 689689, 649649 then this form of number is exactly divisible by 7, 11, 13, 1001.
● If a is divisible by b then ac is also divisible by b.
● If a is divisible by b and b is divisible by c then a is divisible by c.
● If two numbers a & b are divisible by a third number c then (a – b) is also divisible by c.

Successive Division:- successive division, also known as repeated division, involves dividing a number by a sequence of divisors, where quotient from one divisor becomes the dividend for the next. This process continues until a quotient of zero is obtained.

dR×(d×Q + R) = Q(d×Q + R)NumberdR×(d×Q + R) = QdR×(d×Q + R) = QdR×Q ⩽ Rfor least number put Q = 0

Ex:- Successive division of a number by 2 3, 5 and 7 gives remainder 1, 4, 0 and 5 respectively. What will be the sum of remainders if same number is divided by 7, 5, 3 and 2 successively ?
Sol:

2 1×+793 4×+255 0×+57 5×+0159Number
7 22 51595 4 23 1 12 0 1Sum = 9 Answer

(Q). A certain number when successively divided by 4, 5 and 7 leaves remainder 2, 3, 4 respectively. Find such least number
(i) 374     (ii) 514     (iii) 234     (iv) 92
Sol:

4 2×+235 3×+ 47 4×+04 2×+58945 3×+117 4×+1234option (iii)Answer

● Product of n consecutive number is always divisible by n!
Ex: n(n + 1)(n + 2)(n + 3)(n + 4) ⟶ divisible by 5!
(n – 3)(n – 2)(n – 1)(n)(n + 1)(n + 2) ⟶ divisible by 6!

(Q). n² – n is always divisible by which number.
Sol:
(n – 1)(n)(n + 1)
⟶ divisible by 3! = 6      Answer

❓ Frequently Asked Questions on Divisibility

Q1. What is divisibility and why is it important in competitive exams?

Divisibility refers to the property of a number being exactly divided by another number without leaving any remainder. Divisibility rules are shortcut tests that allow you to determine whether a number is divisible by another number without performing actual division. These rules are extremely important in competitive exams because they save time in solving number theory, remainder theorem and simplification problems. Divisibility is a fundamental concept in number theory and is directly connected to factors, multiples and prime numbers.

Q2. What are the divisibility rules for 2, 3, 4 and 5?

For divisibility by 2, the last digit must be even or zero. For divisibility by 3, the sum of all digits must be divisible by 3. For divisibility by 4, the number formed by the last two digits must be divisible by 4. For divisibility by 5, the last digit must be 0 or 5. These four rules are the most commonly tested in SSC CGL, Bank PO, Railway RRB and AMCAT exams. Understanding these rules also helps in solving Remainder Theorem problems faster.

Q3. What is the divisibility rule for 7 and how many methods are there?

There are three methods to check divisibility by 7. Method 1 — take the last digit, multiply by 2, subtract from the rest of the number and repeat until you get a small number. If result is 0 or divisible by 7 the number is divisible by 7. Method 2 — add 5 times the last digit to the rest and check if result is divisible by 7. Method 3 — split number into groups of three from right, calculate alternating sum of groups and check if divisible by 7. All three methods are explained with solved examples on this page. Divisibility by 7 is also useful when solving Factors problems.

Q4. What is the divisibility rule for 11?

A number is divisible by 11 if the difference between the sum of digits at odd positions and the sum of digits at even positions is either 0 or a multiple of 11. For example for the number 36234, sum of digits at odd positions is 3 plus 2 plus 4 equals 9 and sum of digits at even positions is 6 plus 3 equals 9. Difference is 9 minus 9 equals 0 so the number is divisible by 11. This rule is directly related to the Remainder Theorem shortcut for division by 11.

Q5. What is successive division and how is it different from normal division?

Successive division involves dividing a number by a sequence of divisors one after another where the quotient from one division becomes the dividend for the next division. This process continues until the quotient becomes zero. Successive division is used to find the original number when remainders from multiple divisions are given. For example if a number gives remainders 1, 4, 0 and 5 when successively divided by 2, 3, 5 and 7, you can work backwards to find the original number. This concept is closely linked to Remainder Theorem and LCM and HCF.

Q6. What are the general properties of divisibility?

There are several important general properties of divisibility. If a number x is divisible by y then any number divisible by x is also divisible by y and all factors of y. If a number is divisible by two or more co-prime numbers then it is also divisible by their product. If two numbers are both divisible by a third number then their sum and difference are also divisible by that third number. A number written by repeating any digit 6 times is always divisible by 7, 11, 13, 37 and 1001. These properties are useful in solving Number of Zeroes and Factors problems.

Q7. Which competitive exams ask divisibility questions?

Divisibility questions are asked in SSC CGL, SSC CHSL, Bank PO, Bank Clerk, Railway RRB, UPSC CSAT, CAT and all major Indian competitive exams. This topic is also tested in GMAT, GRE, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests worldwide. Divisibility rules for 7, 11 and 13 are particularly popular in SSC CGL and Railway RRB. Successive division problems are common in Bank PO and CAT exams. Related topics to study alongside are Remainder Theorem, Unit Digit and Factors.

Q8. How can I practice divisibility questions for competitive exams?

After understanding all divisibility rules and properties you can practice on our Exercise on Divisibility page which contains a large number of solved practice questions covering all types of divisibility problems asked in SSC CGL, Bank PO, Railway RRB, GMAT, AMCAT, eLitmus and TCS NQT exams. You may also want to study the related topics of Remainder Theorem, Unit Digit, Factors and Number of Zeroes to strengthen your complete number theory preparation. All questions come with detailed step by step solutions and everything is completely free with no registration required.