# Cyclicity

In CAT, MAT and other Competitive examinations like Bank PO, etc. you get questions where you need to find the last digit of numbers raised to large powers. It’s almost impossible to calculate the values of such numbers manually and hence to find digit at their unit’s place. Such problems can be solved using the concept commonly known as **Cyclicity of Numbers**. Here in this post I am explaining in details the concept of cyclicity and how it should be used for solving such problems.

## Finding Last Digit of Any Number Raised to Any Power

“Hello sir , I like your shortcuts very much. But please tell me how to find unit’s digit in numbers like 7 raise to power 205, 19^239. Sir I am pissed off solving these type of problems”- Excerpt of the comment on an earlier post - Shortcut Method for Multiplication

**Cyclicity Explained **

To understand cyclicity let us take a simple example.

Take any two numbers say 43 and 97.

If they are multiplied, the answer is 4171. The last digit of the product is same as the last digit of 3 x 7.

Hence, it is 1.

This concept could be extended to a host of situations. An interesting pattern emerges when we look at the exponents of the numbers. We would find conclusions as given below.

The last digits of the exponents of all numbers have cyclicity i.e. every Nth power of the base shall have the same last digit, if N is the cyclicity of the number. All numbers ending with 2, 3, 7, 8 have a cyclicity of 4.

For instance,

2^1 ends with 2

2^2 ends with 4

2^3 ends with 8

2^4 ends with 6

2^5 end with 2 again.

The same set of the last digits shall be repeated for the subsequent powers. So, if we want to find the last digit of (say) 2^45, divide 45 by 4.

The remainder is 1

So the last digit would be the same as last digit of 2^1, which is 2

Let us take a CAT level example

**The digit in the unit place of the number represented by (7^95 * 3^58) is**

A. 7

B. 0

C. 6

D. 4

**Answer:** A (7)

**Solution**

Cycle of 7 is

7^{ 1}=7

7^{ 2}=49

7 ^{3}= 343

7 ^{4}= 2401

If we divide 95 by 4, the remainder will be 3.

So the last digit of (7)^{95} is equals to the last digit of (7)^{3} i.e. 3.

Cycle of 3 is

3^{1} =3

3^{2 }=9

3^{3}= 27

3^{4}= 81

3^{5}= 243

If we divide 58 by 4, the remainder will be 2. Hence the last digit will be 9.

Therefore, unit’s digit of (7^95 * 3^58) is unit’s digit of product of digit at unit’s place of 7^95 and 3^58 = 3 * 7 = 21. Hence 1 is the answer.

Working out similarly for all other digits we get

CYCLICITY TABLE

1 | 1 |

2 | 4 |

3 | 4 |

4 | 2 |

5 | 1 |

6 | 1 |

7 | 4 |

8 | 4 |

9 | 2 |

10 | 1 |

Using the above table try answering the questions raised in the comment by a QuickerMaths follower, re-posted above.

Ur question of example which is shown above which is …

Cyclicity of 7^95*3^58 ?

Ur answer is 7, which is correct.

But in solution you shows answer is 1. Which is wrong .

So plzz correct it.

Ur question of example which is shown above which is …

Cyclicity of 7^95*3^58 ?

Ur answer is 7, which is correct.

But in solution you shows answer is 1. Which is wrong .

So plzz correct it.

How do you solve this question:

Find the units digit in:

12^(4n+1)+23^(4n+1)+34^(4n+3)-47^(4n-2)

correction in the question:It is 23^(4n+2) and not 23^(4n+1)

6

the difference between a two digit number and sum of its digit is always a multiple of.?

a)6

b)9

c)11

d)1

give reason please.?

Let a no. be 10x+y.

Sum of digits is x+y.

Their difference = 10x +y- x-y

=9x

Which is divisible by 9.

the number is XY then

two digit number is 10X+Y,

sum of digits X+Y,

then differences is 10X+Y-X-Y=9X so the answer is 9

Excelently done…thk uu all

Excellent…..u r nt giving us the solution…u r giving a teacher like mr.india who is nt thr bt do everythng…thank uuuu

Thanks for such a wonderful remark. I feel humbled.

nice trick..thanxx

sir plz tel me how to balance the decimal quikly.

sir plz tel me some tricks about calender clock problems. Plz

Therefore, unit’s digit of (7^95 * 3^58) is unit’s digit of product of digit at unit’s place of 7^95 and 3^58 = 3 * 9 = 27. Hence 7is the answer.

thankyou very much

Sir,

master technique ( cyclicity). I m lucky enough to have access to your lessons. thanks alot.

Can someone explain the table??i can understand others but cant understand the Table..i will be thankful…

Can someone explain the table??i can understand others but cant understand the Table..

infact the table above that you have given above can be simplified by saying that the cyclicity for any number is 4…every digit starts repeating at x^5, where x is the last digit of the number….

great job…

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Thank u buddy!! Great work…….

superb work!!!!!!!!!!!!!!!!! can u send some of them 2my mail.plzzzzz

superb!!!!!!!!!!! work” can u send some more of them to my mail… plz

Thanks for your question.

The answers are as follows:

1) 3

2) 4

3) 5

I downloaded the free e-book. It’s really very exhaustive.

Thanks for the good work.

Can you provide answer for the following questions asked in the e-book

Find out the last digit of

1) 3^57

2) 7^23 x 8^13

3) 235^1000

(1)3

(2)4

(3)5

raman could you show me the whole solution for 235^1000? Coz I dont know how to solve it for a number having more than 1 digit in it. just like 19^239 (mentioned above) has 19, a 2 digit number. Also give me some more examples

1. 3^57

Basically to find unit’s place digit you should know its cyclicity.

=>Cyclicity of 3 =4

Divide the power by cyclicity of the base’s unit’s digit

i.e 57/4 which gives 1 as remainder..now last digit comes from (unit digit of number)^remainder.

i.e (3)^1=3.

Hope this helped you.

Similarly you can solve the other question

235^1000 its unit place digit is 1

Since 1000 z divisible by 4 n the unit digit of 235 z 5 an odd

So as pr formula ans z 1