Problem on cyclicity is a very important topic in the competitions and placement aptitude. Cyclicity aptitude questions are very frequently asked in the all type of examinations even if it is an Engineering entrance or it is a Govt Competition or It is a job placement written test in the aptitude section. So you have to learn how to solve number cyclicity problems easily.

You can’t follow the universal method to solve these type questions in the examination because it may take too much time and in the competition exam there is time limitation. So try these short tricks to solve Problem on number cyclicity.

There are mainly three types of Questions asked in the examination from the Number Cycliclity. and these are given below :

1. Find the unit digit of a^{b
}2. Find the unit digit of a^{b }×c^{d}×e^{f
}3. Find the unit digit of _{a}b^{c}

Let’s find the last digit starting with 2

2^{1} = 2

2^{2 }= 4

2^{3} = 8

2^{4} = 16

2^{5 }= 32

2^{6 }=64

2^{7} = 128

2^{8} = 256

2^{9 }= 512

From the above table you can easily Observe that the last place or unit digit is repeating again again after 4 steps. So now from this series we can evaluate the unit digit of the questions .Let’s take an simple example

**Example 1:** Find the unit digit of 2^{9.
Solution : The power of 2 is 9 and you have to divide the power by 4 and find out the remainder.
9 / 4 = 2 × 4 + 1
Now the question becomes 2 (2×4 ) ×21 .
}Here the remainder is 1 and in the table 1st digit is 2 . So the **Unit digit is 2.
If we solve 2 ^{9 }then we get 512 means or method is correct.^{
}**

Let’s take another example . This time little bit tough.

**Example 2 :** Find the Unit digit of 2^{87 }.

**Solution :** The power of 2 is 87 and you have to divide the power by 4 and find out the remainder.

87 / 4 = 21 × 4 + 3

Now the question becomes **2 ^{(21×4 ) }×2^{3}**

Here the remainder is 3 that means the unit place of 2^{87 }is decided by 3rd item in the table i.e. 8 .

You can try with all questions with the help of this table

No/power | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Cyclicity |

2 | 4 | 8 | 6 | 2 | 4 | 8 | 6 | 4 |

3 | 9 | 7 | 1 | 3 | 9 | 7 | 1 | 4 |

4 | 6 | 4 | 6 | 4 | 6 | 4 | 6 | 2 |

5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 1 |

6 | 6 | 6 | 6 | 6 | 6 | 6 | 6 | 1 |

7 | 9 | 3 | 1 | 7 | 9 | 3 | 1 | 4 |

8 | 4 | 2 | 6 | 8 | 4 | 2 | 6 | 4 |

9 | 1 | 9 | 1 | 9 | 1 | 9 | 1 | 2 |

**Problem on number cyclicity Part 2**

Now move to second type problem (a^{b }×c^{d}×e^{f })

This type of problems are also very simple and you can find the unit digit with the help of above table. Now you are thinking how? . Then let’s see

First find the unit Digit of all the numbers individually means unit digit of a , unit digit of c and unit digit of e. Now multiply the all unit digit. You will get a new number by multiplying and the unit digit of the new number is the unit digit of the question. I think you are not believing to me :O . Let’s take an simple example

**Example 3:** Find the unit digit of 2^{5} × 3^{6} × 6^{4
}**Solution : **Now from the table you can easily calculate the unit digit of all number and that is 2 , 9 and 6 respectively. Now we can find the unit digit by multiplying the unit digit of these tree numbers.

2×9×6 = 108 means the unit digit is ‘8’.

Let’s cross check it by conventional way

2^{5 } = 32

3^{6 }= 729

6^{4} = 1296

Now multiply all these number = 32 × 729 × 1296 = 30233088

Means our method is correct.

**Problem on number cyclicity Part 3**

Now come to the third and most important part of this type problem. This type of Questions are mostly asked in the Engineering Entrance examinations. So let’s take an example how to solve these type of problem on number cyclicity.

**Example 4 :**Find the units place digit of _{2}23^{34
}**Solution :**The cyclicity for 2 at units place digit is 4. So we can find out the unit digit by the first step. But in slightly different manner.

Now we have to find the remainder when exponent of 2 is divided by 4, that is the remainder when 23^{34} is divided by 4.

Remainder of 23^{34}/4 = Remainder of (24 – 1)^{34}/4

Using the binomial theorem, we can see that there is only one term in the expansion of (24 – 1)^{34} which is not divisible by 4.

The term is 13^{4}/4

Remainder of 13^{4}/4 = 1

Now from the above table we can easily find the unit digit and i.e. 2 .

Now you can easily find the unit digit by the cyclic method.

**Also Read**: **How to solve Time and Work Problem **

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