In most of the quantitative aptitude tests you might have faced questions like what’s the units digit of **9 ^{290}** or

**23**etc. Needless to say, it’s practically impossible to do the direct calculation to find the answer considering the time constraint of quantitative aptitude exams. But if you notice, you can clearly see that there exists a pattern when you apply same mathematical operation continuously on a number. This is called cyclicity of numbers. In this paper we will have a look at top Cyclicity of Numbers aptitude questions, concepts and shortcuts on how to solve units digit problems quickly using concepts of number cyclicity

^{512}Problems based on cyclicity of numbers are common in most **MBA** entrance examinations like **CAT, MAT, XAT, GMAT** and other competitive exams like BANK tests. Questions related to number cyclicity have also started appearing in career recruitment tests or placement tests conducted by various companies. Since last few years **TCS** has redesigned their quantitative aptitude test patterns which include questions based on cyclicity of numbers. Questions like **' find the units digit in the result of a^{b} '** are difficult to solve if you do them manually.

Instead if you apply theory of cyclicity of numbers, these questions does not take much of your time to solve them. Remember time is a constraint all competitive exams. So picking up concepts of cyclicity of numbers is definitely a bonus. Take the online test provided at the end to check your understanding level once you complete this tutorial on number cyclicity.

You can also try

- Tricks to crack aptitude questions on Numbers
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## Important Concepts and Shortcuts of Cyclicity of Numbers

**Types of questions based on cyclicity of numbers**

There are mainly 3 categories of questions which fall under cyclicity of numbers, that include

- How to find units digit of a
^{b} - How to find units digit of a
^{b}* c^{d}* e^{f} - How to find units digit of a
^{bc}

## Find units digit of a^{b}

**a**, units place digit of the result depends on units place digit of

^{b}*and the divisibility of power*

**a****.**

*b*Consider powers of 2

As we know,

^{1}=2

2

^{2}= 4

2

^{3}= 8

2

^{4}= 16

2

^{5}= 32

2

^{6}= 64

2

^{7}= 128.. and so on

*What do you observe here?*We can see that the units place digit for powers of 2 repeat in an order: 2, 4, 8, 6. So the "cyclicity" of number 2 is 4 (

*that means the pattern repeats after 4 occurrences*) and the cycle pattern is 2, 4, 8, 6. From this you can see that to find the units place digit of powers of 2, you have to divide the exponent by 4.

Let's check the validity of above formula with an example.

**Example**

- Find the units place digit of 2
^{99}?

^{99}is the 3rd item in the cycle which is 8.

**Shortcuts to solve problems related to units place digit of a**

^{b}**Case 1: If b is a multiple of 4**- If
**a**is an even number, ie: 2, 4, 6 or 8 then the units place digit is 6 - If
**a**is an odd number, ie: 1, 3, 7 or 9 then the units place digit is 1 **Case 2: If b is not a multiple of 4**- Let
**r**be the reminder when**b**is divided by 4, then units place of**a**will be equal to units place of^{b}**a**^{r}

**Here we have captured the cyclicity of numbers upto 9 in the below table.**

Number | ^1 | ^2 | ^3 | ^4 | Cyclicity |
---|---|---|---|---|---|

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

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

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

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

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

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

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

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

## Find units digit for numbers of the form a^{b} * c^{d}

- First find the unit digit of
**a**and^{b}**c**separately. Let the answers be^{d}**x**and**y** - Then unit digit of
**a***^{b}**c**= units digit of^{d}**x * y**

## Find units digit of a^{bc}

**Case 1:**If cyclicity of units place digit of a is 4 then we have to divide the exponent of a by 4 and find out the remainder. Depending on the value of remainder we can apply the general rule of cyclicity given above and reach the solution.**Case 2:**If cyclicity of units place digit of a is 2, only extra information we need to find is if the exponent will be even or odd. Then we can apply the general rule of cyclicity given above and reach the solution.

**Example 1:**

- Find the units place digit of 2
^{4344}

- Here cyclicity of units place digit is 4 (Units place digit is 2, from the above table we can see the cyclicity of 2 is 4). Hence case 1 is applicable.
- Now we have to find the remainder when exponent of 2 is divided by 4, that is the remainder when 43
^{44}is divided by 4. - Remainder of 43
^{44}/4 = Remainder of (44 – 1)^{44}/4 - Using the binomial theorem, (as explained in number system tutorial) we can see that there is only one term in the expansion of (44 – 1)
^{44}which is not divisible by 4. - The term is 1
^{44}/4 - Remainder of 1
^{44}/4 = 1 - Now we can apply the general rules of cyclicity, (since reminder is 1, case 2 of general rule of cyclicity is applicable) which says, units place of 2
^{4344}= units place of 2^{1}= 2.

**Example 2:**

- Find the units place digit of 29
^{3945}

- Here cyclicity of units place digit of a, that is cyclicity of 9 is 2. Hence case 2 is applicable
- Next step is we need to find if 39
^{45}is even or odd - 39
^{45}will always result in an odd number, because we are multiplying an odd number 39, odd number of times (45). - Now we can apply the general rule of cyclicity described in the first section. General rule says, first check if exponent is divisible by 4. Since the exponent here is odd, it's not divisible by 4.
- Again according to general rule, if exponent is not divisible by 4, find the reminder when exponent is divided by 4.
- Since the exponent is odd here, the possible reminders when it's divided by 4 are 1 and 3.
- This means, the units place digit of number will be 1st or 3rd element in the cyclicity of units place digit of a.

In this case 1st and 3rd elements of 9 (units place digit of 29) is 9.

Hence units place digit of 229^{3945} is 9.