Sohel Sahoo : Hello Guys,

In this  lesson you are going to learn about an important concept of Cyclicity Theorem Or Pattern Theorem . Cyclicity  theorem gives a pattern of remainders, which can be used to solve questions based on remainders. This concept utilizes the fact that remainders repeat themselves after a certain interval when divided by a number.

CYCLICITY THEOREM OR PATTERN THEOREM:-

This theorem states that if we divide aⁿ by d the remainder can take any value from 0 to d-1. If we keep on increasing the value of n, the remainders are cyclical in nature. The pattern of the remainders would repeat.

Let us understand the concept of repetition  in better with the help of an example.

Method-1 [Using previous basic concept of solving]

  Q.  Find the remainder when 37¹⁰⁰ is divided by 5 . 

Sol:-  (37)¹⁰⁰/5  [Here 37>5;so 1^st divide this & obtain remainder]

 = 2¹⁰⁰/5

=  (2²)⁵⁰/5  [As 2² = 4 & 5-1=4]

=  4⁵⁰/5

=  (-1)⁵⁰

= 1 (Ans)

Method-2 [ Using Cyclicity theorem ]

 Q.  Find the remainder when 37¹⁰⁰ is divided by 5 . 

Sol:-  (37)¹⁰⁰/5  [Here 37>5;so 1^st divide this & obtain remainder]

 = 2¹⁰⁰/5

Now to show that it forms a pattern we could generalize by  2^N/5 ;Where, N=1,2,3,4,5....etc.natural numbers.

 Look that,   2¹/5 = 2

                      2²/5 = -1 / 4

                     2³/5 = 3

                     2⁴/5 = 1

Clearly,  2⁵/5 =  2 x 2⁴/5 =2x1 = 2

So, the cycle  2^N/5 starts repetiting after 4 powers as above.Hence we have to just divide the power 100 by 4 & find the remainder and the value of  2^Remainder/5 will be the required result i.e. remainder obtained by dividing 37¹⁰⁰ by 5.

Now, R=[100/4]=0 i.e. it is completely divisible by 4.Remainder can be treated as 4.so, answer is 1 (Ans)

NOTE:-

• While determining pattern or cycle of remainders if simply the remainders starts repetiting after a interval or 1 appears;then after that power the remainder will follow the pattern i.e. repetition starts.  [Imp. shortcut trick]

• In finding remainders if -1 appears then after twice(2 times) steps repitition will start as it would become 1 by multiplying itself. [Imp. shortcut trick]                                                       e.g. if -1 appears in 3 steps then repetition will start in 6 steps.

•  The power of a number after which repetition starts is called cyclicity of that number.If the remainder is 0 then it can be treated as cyclicity.
• After finding cyclicity in a^N simply get R= [N/cyclicity] and find a^R . This will be the required remainder which also reduce the steps & save time .

  Q.  Find the remainder when 11⁷⁷ is divided by 7 . 

^Sol:- (11)⁷⁷/7  [Here 11>7;so 1^st divide this & obtain remainder]

 = 4⁷⁷/7

=  (2²)⁷⁷/7 

=  2¹⁵⁴/7

= 2 x  (2³)⁵¹/7  [As 2³ = 8 & 7+1=8]

= 2 x 1⁵¹

= 2 x 1

=  2  (Ans)

# [ Using Cyclicity theorem ]

Sol:-   (11)⁷⁷/7  [Here 11>7;so 1^st divide this & obtain remainder]

        = 4⁷⁷/7

Now to show that it forms a pattern we could generalize by  4^N/7 ;Where, N=1,2,3,4,5....etc.natural numbers.

 Look that,   4¹/7 = 4

                      4²/7 = 2

                     4³/7 = 1

     Clearly, repetition starts after 3 powers. Hence cyclicity is 3 .

Now, R [77/3] = 2

So, required remainder will be 4²/7 = 2 (Ans)

  Q.  Find the remainder when 59²⁸ is divided by 7 . 

^Sol:-  (59)²⁸/7  [Here 59>7;so 1^st divide this & obtain remainder]

        = 3²⁸/7

      Now to show that it forms a pattern we could generalize by  3^N/7 ;Where,     N=1,2,3,4,5....etc.natural numbers.

     Look that,   3¹/7 = 3

                         3²/7 = 2

                         3³/7 = -1 / 6

                        3⁴/7 = 4

                         3⁵/7 = 5

                         3⁶/7 = 1

Here, -1 appears in 3 steps so repetition will start in 2 x 3 = 6 steps. Hence cyclicity is 6 .

Now, R [ 28/6 ] = 4

So, required remainder is  3⁴/7 = 4 (Ans)

Let’s see a difference between the three expressions.

⟨2²⟩³ = 2⁶ 

2²^{3 =} 2⁸

2³^{2 =} 2⁹

Clerly, 2^{6 <} 2^{8  <}  2⁹ 

          ^{=> ⟨2²⟩³ <2²3  <} 2³² 

[ When there is a bracket between two powers then simply multiply the powers.But when the powers are written in raised form then solve it from top as shown above]

  Q.  Find the remainder when 32³²³² is divided by 7 . 

^Sol:-  32³²³²/7 = 32^N/7 ; where N = 32³²

Now, 32^N/7 = 4^N/7  [Here 32>7;so 1^st divide this & obtain remainder]

  Applying cyclicity theorem,

                                   4¹/7 = 4

                                            4²/7 = 2

                                            4³/7 = 1

     Clearly, repetition starts after 3 powers. Hence cyclicity is 3 .

Now,  R[N/3] = R [32³² /3 ] = (-1)³²  = 1

So, required remainder is  4¹/7 = 4 (Ans)

Note:-   32³² /3 = 32 x 32 x 32 ...... x 32 (32 times) /7

            =  (-1) x (-1) x (-1) x......x (-1) (32 times) / 7

            = (-1)³² 

            = 1

  Q.  Find the remainder when 35²³²³ is divided by 16 . 

Sol:-  35²³²³ /16 

       =  3²³²³ /16  [Here 35>16;so 1^st divide this & obtain remainder]

           =  3^N  /16

          Applying cyclicity theorem,

                                   3¹/16 = 3

                                            3²/16 = 7

                                            3³/16 = 11

                                             3⁴/16 = 1

     Clearly, repetition starts after 4 powers. Hence cyclicity is 4 .

Now,  R[N/4] = R [23²³ /4 ] = (-1)²³  = -1 = 4-1 = 3

So, required remainder is    3³/16 = 11 (Ans)

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