This brilliant article was contributed to Total Gadha by Ashish Tyagi, a regular TGite. I do not think there can be easier or more lucid method of finding the last two digits of a number. Kudos to Ashish for this article. We will also welcome similar contributions from our other users. Do you have some useful or life-saving funda that you would like to share? Send it to us and we will publish it with your name- Total Gadha

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 }**× (30) +**

^{785}**C**

^{786}**× 1**

_{2}**× 30**

^{784}**+ ..., 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:

Here are some more examples:

**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}**× 61**

^{456}**will be the last two digits of 01 × 21 = 21**

^{567}**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}**× 33**

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

^{2}**. By the previous method, the last two digits of 21**

^{72}**= 41 (tens digit = 2 × 2 = 4, unit digit = 1)**

^{72}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

**× 87**

^{472}**= (87**

^{2}**)**

^{4}**× 87**

^{118}**= (69 × 69)**

^{2}**× 69 (The last two digits of 87**

^{118}**are 69) = 61**

^{2}**× 69 = 81 × 69 = 89**

^{118}If you understood the method then try your hands on these questions:

**Find the last two digits of:**

1. 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}**× 2**

^{54}**= (24)**

^{3}**(24 raised to an even power) × 2**

^{54}**= 76 × 8 = 08**

^{3}(**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 × 2**

^{n}**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}**× 2**

^{141}**= 24**

^{6}**(24 raised to odd power) × 64 = 24 × 64 = 36**

^{141}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}**× 3**

^{586}**= (2**

^{586}**)**

^{10}**× 2**

^{58}**× 31**

^{6}**= 76 × 64 × 81 = 84**

^{586}**Find the last two digits of 54 ^{380}.**

54** ^{380}** = (2 × 3

**)**

^{3}**= 2**

^{380}**× 3**

^{380}**= (2**

^{1140}**)**

^{10}**× (3**

^{38}**)**

^{4}**= 76 × 81**

^{285}**= 76 × 01 = 76.**

^{285}**Find the last two digits of 56 ^{283}.**

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

**× 7)**

^{3}**= 2**

^{283}**× 7**

^{849}**= (2**

^{283}**)**

^{10}**× 2**

^{84}**× (7**

^{9}**)**

^{4}**× 7**

^{70}**= 76 × 12 × (01)**

^{3}**× 43 = 16**

^{70}**Find the last two digits of 78 ^{379}.**

78** ^{379}** = (2 × 39)

**= 2**

^{379}**× 39**

^{379}**= (2**

^{379}**)**

^{10}**× 2**

^{37}**× (39**

^{9}**)**

^{2}**× 39 = 24 × 12 × 81 × 39 = 92**

^{189}Now try to find the last two digits of

1. 34^{576}

2. 28^{287}

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