How To Solve Divisibility Questions Quickly

April 26, 2020

How To Solve Divisibility Questions Quickly & Definition

The capacity of a dividend to be exactly divided by a given number is termed as divisibility.

One whole number is divisible by another if, after dividing, the remainder is zero.
If the whole number is divisible by another number than the second number is factor of 1st number.

Read Also – Formulas to solve divisibility questions

How To Solve Divisibility Questions Quickly

Type 1: Find the largest or smallest number

Question 1.

What smallest number should be added to 1056 so that the number is completely divisible by 23?

Options

Correct option:1

Solution

On dividing 1056 by 23 we get remainder as 21

Required number = 23 – 21 = 2

1056 + 2 = 1058

1058 is completely divisible by 23 leaving reminder as 0.

Question 2.

In the given numbers, if first digit of each number is replaced by second digit, second digit is replaced by third digit, and third digit is replaced by first digit. Find out the second lowest number?
456 137 564 238 625

Options

Correct option:1

Solution

456 137 564 238 625

On replacing the position as per the instructions given in the question, the number become

564 371 645 328 256

Therefore, the second lowest number is 238

Question 3.

Find the largest 4 digit number which is exactly divisible by 88?

Options

9955
9988
9944
9088

Correct option:3

Solution

Largest 4 digit number is 9999

On dividing 9999 by 88, remainder = 55

Required number = 9999 – 55 = 9944

Type 2: Which of the following numbers is/or not divisible by given number.

Question 1.

Which of these numbers is not divisible by 5?

Options

345675
234565
230050
345601

Correct option:4

Solution

A number is exactly divisible by 5 if it has the digits 0 or 5 at one’s place. Therefore, only option D is not divisible.

Question 2.

Which of these numbers is not divisible by 7?

Options

875
4143
1470
5488

Correct option:2

Solution

For 4143

Remove 3 from the number and double it = 6
Remaining number is 414, now subtract 414 from 6 = 414 – 6 = 408.
Repeat the process, We have last digit as 8, double = 16
Remaining number is 40, now subtract 40 from 16 = 40 – 16 = 24.
As 24 is not divisible by 7, hence the number 414 is not divisible by 7.

Question 3.

How many numbers between 300 and 900 are divisible by 4, 5 and 6?

Options

Correct option:1

Solution

Solution: First find the LCM of 4, 5, and 6 = 60

Now, on dividing 900 by 60 we get quotient as 15

On dividing 300 by 60 we get quotient as 5

Therefore the required number is 15 – 5 = 10

Type 3: How To Solve Divisibility Questions Quickly.
Find the remainder

Question 1.

On dividing a number by 60, we get 159 as quotient and 0 as remainder. On dividing the same number by 50, what will be the remainder?

Options

Correct option:1

Solution

Number = 159 * 60 + 0 = 9540

On dividing 9540 by 50 we get remainder as 40

Question 2.

What is the remainder when (2p + 2)2 is divided by 4 and ‘p’ is an integer?

Options

1
2
3
None of the above

Correct option:4

Solution

On expanding (2p + 2)² = 4p² + 8p + 4

Now, take 4 common, then we get 4(p² + 2p + 1)

Let the value of p be 1

4 (12 + 2*1 + 1) = 16

Hence, 16 is divisible by 4 and the remainder will be 0

Question 3.

Find the remainder when 4⁸⁷⁵is divided by 17.

Options

Correct option:4

Solution

Rem [4⁸⁷⁵ / 17]
= Rem[4*4⁸⁷⁴ / 17]
= Rem [4* 16⁴³⁷ / 17]

; as when 4 * (17-1)⁴³⁷ is divided by 17, it will give remainder of 4 * (-1)
= Rem [4*(-1) / 17]

= Rem [-4 / 17]
= 13

Find Remainders of Large Powers Quickly

Ever wondered how you can calculate remainder of large powers quickly ? Like –

25^26²⁷when divided by 3 what would be result?

You would need to learn the following –

Basic Rules
Common Factor
Co Primes

Will be added later –

Fermat Theorem
Wilson Theorem
Euler Theorem

Basic Rules

Rule 0

Rem[(x^y)/d] = Rem[r^y/d]

where r is remainder when x divided by d

Example

Rem[(31³²)/3]
= Rem[1³²/3] (as remainder is 1 when 31 is divided by 3)
= 1

Rem[(29²⁹)/3] = Rem[(-1)³²/3] = -1 = +2

The above -1 can be written as 2 as -1 remained in case of 3 is nothing but + 2 remainder.

We didn’t do 2 power 32 as it would have taken additional steps and would’ve been lengthy

Rem[(29³⁰)/3] = Rem[(-1)³⁰/3] = 1

Rule 1

Rem[(a*b*c)/d] = Rem[a/d] * Rem[b/d] * Rem[c/d]

Example

Rem[(30*31*32)/7] = Rem[30/7] * Rem[31/7] * Rem[32/7]
Rem[2] * Rem[3] * Rem[4]
Rem[2*3*4] => Rem[24/7]
3

Rule 2

Rem[(a+b+c)/d] = Rem[a/d] + Rem[b/d] + Rem[c/d]

Example

Rem[(30+31+32)/7] = Rem[30/7] + Rem[31/7] + Rem[32/7]
Rem[2] + Rem[3] + Rem[4]
Rem[2+3+4] => Rem[9/7]
2

Applications of above rules

Rem[(30³⁰)/7]

Solution

Rem[(30³⁰)/7] = Rem[(30)/7] * Rem[(30)/7] * Rem[(30)/7] * Rem[(30)/7] …….. (30 times)
Rem[(30³⁰)/7] = 2 * 2 * 2 ….. (30 times) = 2³⁰
Rem[(30³⁰)/7] = 2³⁰= (2⁶)⁵= (64)⁵
Rem[(30³⁰)/7] = Rem[64/7])⁵ => 1⁵
Rem[(30³⁰)/7] = 1

Rem[(30^31³²)/7]

Solution

Rem[(30^31³²)/7] = Rem[(2^31³²)/7] (as Rem[30/7] = 2)

Hypothesis

We know, 2³will give us remainder 1 when divided by 7 (Rem[8/7] = 1)
Thus, for any K > 0, 2^3Kwill also give us remainder 1 when divided by 7

Futhemore,

2^3K+1will give us remainder 2
1. 2^3K+1can be written as (2^3K)*2 => Rem[2^3K/7] * 2
2. 1* 2 = 2

Similarly,

2^3K+2will give us remainder 4
1. 2^3K+2can be written as (2^3K)*2*2 => Rem[2^3K/7] * 2 * 2
2. 1* 2 * 2 = 4

Now, we need to write 31³²in 3K, 3K + 1, 3K + 2 format by doing follows –

Rem[31³²/3] = Rem[1³²/3] = 1

Thus, can be written in format 3k+1

Rem[(30^31³²)/7] = Rem[(2^31³²)/7] = Rem[(2^3K+1)/7] = 2

Thus, the remainder will be 2

Common Factor

Rem[X/Y] = Rem[kx/ky] = k *resultOf(Rem[x/y])

Example

Rem[(4¹⁵)/28] = Rem[(4*4¹⁴)/(4*7)] = 4* Rem[(4¹⁴)/(7)]
4 * resultOf(Rem[(4*4*4¹²)/(7)])
4 * resultOf(Rem[(4)/(7)] * Rem[(4)/(7)] * Rem[(4¹²)/(7)])
4 * resultOf(4 *4 * Rem[(64⁴)/(7)])
4 * resultOf(4 *4 * Rem[(1⁴)/(7)])
4 * resultOf(Rem[(4*4*1)/7])
4 * resultOf(Rem[(16)/7])
4 * resultOf(2])
8

Example

Rem[(5¹⁶)/35] = Rem[(5*5¹⁵)/(5*7)] = 5* Rem[(5¹⁵)/(7)]
5 * resultOf(Rem[(125⁵)/(7)])
5 * resultOf(Rem[(6⁵)/(7)])
5 * resultOf(Rem[(6*36*36)/(7)])
5 * resultOf(Rem[(6)/(7)] * Rem[(36)/(7)] * Rem[(36)/(7)])
5 * resultOf(6 *1 * 1)
30

Co Primes

When trying to find out the remainder, if the divisor can be broken down into smaller co-prime factors; then

Rem[M/N] = Rem[M/(a*b)]

HCF(a, b) = 1 (then only these are co-primes)

Let, Rem[M/a] = r₁ & Rem[M/b] = r₂

Rem[M/N] = axr₂ + byr₁

Such that ax + bx = 1

Example

Rem[(7¹⁵)/15]
Rem[(7¹⁵)/(3*5)]

a = 3 & b = 5

Rem[(7¹⁵)/(3)] = 1 &
Rem[(7¹⁵)/(5)] = Rem[(2¹⁵)/(5)] = 3

r_{2 =}3, r₁= 2

axr₂ + byr₁= 3*x*3 + 5*y*1

Such that, ax + by = 1 | 3x + 5y = 1

Valid values are x = -3 and y = 2

Thus final answer will be: 3*(-3)*3 + 5*2*1 = – 27 + 10 = 17

Fermat Theorem

If p is a prime, and HCF (a, p) = 1 (a and p are co-primes), then Rem[a^p-1/p] = 1

Example

Rem[2³⁴⁵/11]
Rem[((2¹⁰)³⁴(2))/11]
Rem[((2¹⁰)³⁴)/11] * Rem[(2)⁵)/11]
p = 11 and p – 1 = 10 so using fermats theorem
Rem[((1)³⁴)/11] * Rem[(2)⁵)/11]
Rem[((1)³⁴)/11] * Rem[(2)⁵)/11]
Rem[1/11] * Rem[32)/11]
1 * 10
10

How To Solve Divisibility Questions Quickly

How To Solve Divisibility Questions Quickly & Definition

How To Solve Divisibility Questions Quickly

Type 1: Find the largest or smallest number

Question 1.

Correct option:1

Solution

Question 2.

Correct option:1

Solution

Question 3.

Correct option:3

Solution

Type 2: Which of the following numbers is/or not divisible by given number.

Question 1.

Correct option:4

Solution

Question 2.

Correct option:2

Solution

Question 3.

Correct option:1

Solution

Type 3: How To Solve Divisibility Questions Quickly.Find the remainder

Question 1.

Correct option:1

Solution

Question 2.

Correct option:4

Solution

Question 3.

Correct option:4

Solution

Find Remainders of Large Powers Quickly

Basic Rules

Rule 0

Example

Rule 1

Example

Rule 2

Example

Applications of above rules

Solution

Solution

Hypothesis

Common Factor

Example

Example

Co Primes

Example

Fermat Theorem

Example

Type 3: How To Solve Divisibility Questions Quickly.
Find the remainder