Divisibility Rules and Remainders for Placement Aptitude

Divisibility Rules and Remainders for Placement Aptitude

Divisibility and remainder problems are staple questions in aptitude for placements. Companies like TCS and Cognizant frequently test whether candidates can quickly determine divisibility and find remainders without lengthy calculations. Mastering these rules saves valuable time in time-pressured placement exams.

Divisibility Rules

A number is divisible by 2 if its last digit is even. Divisible by 3 if the digit sum is divisible by 3. Divisible by 4 if the last two digits form a number divisible by 4. Divisible by 5 if the last digit is 0 or 5. Divisible by 6 if divisible by both 2 and 3. Divisible by 9 if the digit sum is divisible by 9. Divisible by 11 if the alternating digit sum (odd positions minus even positions) is 0 or a multiple of 11.

Unit Digit Cyclicity

Unit digits of powers of 2 follow the cycle 2, 4, 8, 6 with period 4. Powers of 3 cycle as 3, 9, 7, 1 with period 4. Powers of 7 cycle as 7, 9, 3, 1 with period 4. Powers of 5 always end in 5. Powers of 6 always end in 6. To find the unit digit of a large power, divide the exponent by the cycle length and use the remainder as the position in the cycle.

Solved Example

What is the remainder when 2 to the power 50 is divided by 7? Powers of 2 modulo 7 cycle as 2, 4, 1 with period 3. Since 50 equals 3 times 16 plus 2, the answer equals 2 to the power 2 modulo 7, which is 4. The remainder is 4.

Interview Tips

In online aptitude tests with time pressure, divisibility rules can eliminate answer choices in seconds without any calculation. Practice identifying cyclicity periods of common bases. Many placement tests include at least one remainder problem with a large exponent — recognizing the cyclicity pattern is the fastest path to the correct answer every time.