## Last two digits of a number ( Method of finding Last two digits)

I am dividing this method into four parts and we will discuss each part one by one:

a. Last two digits of numbers which end in one

b. Last two digits of numbers which end in 3, 7 and 9

c. Last two digits of numbers which end in 2

d. Last two digits of numbers which end in 4, 6 and 8

Before we start, let me mention binomial theorem in brief as we will need it for our calculations.

**Last two digits of numbers ending in 1**

Let’s start with an example.

**What are the last two digits of 31 ^{786}?**

Solution: 31** ^{786}** = (30 + 1)

**=**

^{786}**C**

^{786}** 1**

_{0 }×**+**

^{786}**C**

^{786}** 1**

_{1 }×

^{785}**×** (30) +

**C**

^{786}** 1**

_{2}×

^{784}**×** 30

**+ …, Note that all the terms after the second term will end in two or more zeroes. The first two terms are**

^{2}**C**

^{786}** 1**

_{0 }×**and**

^{786}**C**

^{786}** 1**

_{1 }×**× (30). Now, the second term will end with one zero and the tens digit of the second term will be the product of 786 and 3 i.e. 8. Therefore, the last two digits of the second term will be 80. The last digit of the first term is 1. So the last two digits of 31**

^{785}**are 81.**

^{786}Now, here is the shortcut:

**Find the last two digits of 41 ^{2789}**

In no time at all you can calculate the answer to be 61 (4 **×** 9 = 36). Therefore, 6 will be the tens digit and one will be the units digit)

**Find the last two digits of 71 ^{56747}**

Last two digits will be 91 (7 **×** 7 gives 9 and 1 as units digit)

Now try to get the answer to this question within 10 s:

**Find the last two digits of 51 ^{456} **

**×**

**61**

^{567}The last two digits of 51** ^{456}** will be 01 and the last two digits of 61

**will be 21. Therefore, the last two digits of 51**

^{567}

^{456}**×** 61

**will be the last two digits of 01**

^{567}**×**21 = 21

**Last two digits of numbers ending in 3, 7 or 9**

**Find the last two digits of 19 ^{266}.**

19** ^{266}** = (19

**)**

^{2}**. Now, 19**

^{133}**ends in 61 (19**

^{2}**= 361) therefore, we need to find the last two digits of (61)**

^{2}**.**

^{133}Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 **×**3 = 18, so the tens digit will be 8 and last digit will be 1)

**Find the last two digits of 33 ^{288}.**

33** ^{288}** = (33

**)**

^{4}**. Now 33**

^{72}**ends in 21 (33**

^{4}**= 33**

^{4}

^{2}**×**33

**= 1089**

^{2}**×**1089 = xxxxx21) therefore, we need to find the last two digits of 21

**. By the previous method, the last two digits of 21**

^{72}**= 41 (tens digit = 2**

^{72}**×**2 = 4, unit digit = 1)

So here’s the rule for finding the last two digits of numbers ending in 3, 7 and 9:

Now try the method with a number ending in 7:

**Find the last two digits of 87 ^{474}.**

87** ^{474}** = 87

^{472}**×**87

**= (87**

^{2}**)**

^{4}

^{118}**×**87

**= (69**

^{2}**×**69)

^{118}**×**69 (The last two digits of 87

**are 69) = 61**

^{2}

^{118}**×** 69 = 81

**×** 69 = 89

If you understood the method then try your hands on these questions:

**Find the last two digits of:**

- 27
^{456}

2. 79^{83}

3. 583^{512}

**Last two digits of numbers ending in 2, 4, 6 or 8**

There is only one even two-digit number which always ends in itself (last two digits) – 76 i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as last two digits for even numbers. We know that 24** ^{2}** ends in 76 and 2

**ends in 24. Also, 24 raised to an even power**

^{10}**always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, 24**

**will end in 76 and 24**

^{34}**will end in 24.**

^{53}Let’s apply this funda:

**Find the last two digits of 2 ^{543}.**

2** ^{543}** = (2

**)**

^{10}

^{54}**×**2

**= (24)**

^{3}**(24 raised to an even power)**

^{54}**×**2

**= 76**

^{3}**×**8 = 08

(**NOTE: **Here if you need to multiply 76 with 2** ^{n}**, then you can straightaway write the last two digits of 2

**because when 76 is multiplied with 2**

^{n}**the last two digits remain the same as the last two digits of 2**

^{n}**. Therefore, the last two digits of 76**

^{n}**×**2

**will be the last two digits of 2**

^{7}**= 28)**

^{7}Same method we can use for any number which is of the form 2** ^{n}**. Here is an example:

**Find the last two digits of 64 ^{236}.**

64** ^{236}** = (2

**)**

^{6}**= 2**

^{236}**= (2**

^{1416}**)**

^{10}

^{141}**×**2

**= 24**

^{6}**(24 raised to odd power)**

^{141}**×**64 = 24

**×**64 = 36

Now those numbers which are not in the form of 2n can be broken down into the form 2n ´ odd number. We can find the last two digits of both the parts separately.

Here are some examples:

**Find the last two digits of 62 ^{586}.**

62** ^{586}** = (2

**×**31)

**= 2**

^{586}

^{586}**×**3

**= (2**

^{586}**)**

^{10}

^{58}**×** 2

^{6}**×**31

**= 76**

^{586}**×**64

**×**81 = 84

**Find the last two digits of 54 ^{380}.**

54** ^{380}** = (2

**×**3

**)**

^{3}**= 2**

^{380}

^{380}**×**3

**= (2**

^{1140}**)**

^{10}

^{38}**×**(3

**)**

^{4}**= 76**

^{285}**×**81

**= 76**

^{285}**×**01 = 76.

**Find the last two digits of 56 ^{283}.**

56** ^{283}** = (2

^{3}**×**7)

**= 2**

^{283}

^{849}**×**7

**= (2**

^{283}**)**

^{10}

^{84}**×**2

^{9}**×**(7

**)**

^{4}

^{70}**×**7

**= 76**

^{3}**×**12

**×**(01)

^{70}**×**43 = 16

**Find the last two digits of 78 ^{379}.**

78** ^{379}** = (2

**×**39)

**= 2**

^{379}

^{379}**×**39

**= (2**

^{379}**)**

^{10}

^{37}**×**2

^{9}**×**(39

**)**

^{2}

^{189}**×**39 = 24

**×**12

**×**81

**×**39 = 92

Now try to find the last two digits of

1. 34^{576}

2. 28^{287}

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### 2 comments

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Thnks for these types of Question. Really helps a lot.

It is very helpfull for competitive exams.