Cyclicity

Cyclicity

In CAT and other MBA entrance examinations you get questions based on cyclicity every year. Here we are explaining in details the concept of cyclicity and how it should be used for solving problems.

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

(3) The digit in the unit place of the number represented by (7⁹⁵ – 3⁵⁸) is

A. 7
B. 0
C. 6
D. 4

Answer: D (4)

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 7 is

3¹ =3

3² =9

3³= 27

3⁴= 81

3⁵= 243

If we divide 58 by 4, the remainder will be 2.

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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

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