Cyclicity ............[competitive. exam prep.... topic 10]

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

31 =3

32 =9

33= 27

34= 81

35= 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