How To Solve Divisibility Questions Quickly

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.

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

  1. 2
  2. 0
  3. 1
  4. 3

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

  1. 238
  2. 625
  3. 456
  4. 137

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

  1. 9955
  2. 9988
  3. 9944
  4. 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

  1. 345675
  2. 234565
  3. 230050
  4. 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

  1. 875
  2. 4143
  3. 1470
  4. 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

  1. 10
  2. 13
  3. 14
  4. 15

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

  1. 40
  2. 10
  3. 20
  4. 30

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. 1
  2. 2
  3. 3
  4. 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 4875 is divided by 17.

Options

  1. 19
  2. 12
  3. 21
  4. 13

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 –

252627when 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
How to find remainders of large numbers quickly

Basic Rules

Rule 0

Rem[(xy)/d] = Rem[ry/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

or

  • 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

or

  • 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[(303132)/7]

Solution

  • Rem[(303132)/7] = Rem[(23132)/7] (as Rem[30/7] = 2)

Hypothesis

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

Futhemore,

  1. 23K+1 will give us remainder 2
    1. 23K+1 can be written as (23K)*2 => Rem[23K/7] * 2
    2. 1* 2 = 2

Similarly,

  1. 23K+2 will give us remainder 4
    1. 23K+2 can be written as (23K)*2*2 => Rem[23K/7] * 2 * 2
    2. 1* 2 * 2 = 4

Now, we need to write 3132 in 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[(303132)/7] = Rem[(23132)/7] = Rem[(23K+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

r2 = 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[ap-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