Last Two Digits of a Number

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 31⁷⁸⁶?

Solution: 31⁷⁸⁶ = (30 + 1)⁷⁸⁶ = ⁷⁸⁶C₀ × 1⁷⁸⁶ + ⁷⁸⁶C₁ × 1⁷⁸⁵ × (30) + ⁷⁸⁶C₂ × 1⁷⁸⁴ × 30² + ..., Note that all the terms after the second term will end in two or more zeroes. The first two terms are ⁷⁸⁶C₀ × 1⁷⁸⁶ and ⁷⁸⁶C₁ × 1⁷⁸⁵ × (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 31⁷⁸⁶ are 81.

Now, here is the shortcut:

Here are some more examples:

Find the last two digits of 41²⁷⁸⁹

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 71⁵⁶⁷⁴⁷

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 51⁴⁵⁶ × 61⁵⁶⁷

The last two digits of 51⁴⁵⁶ will be 01 and the last two digits of 61⁵⁶⁷ will be 21. Therefore, the last two digits of 51⁴⁵⁶ × 61⁵⁶⁷ will be the last two digits of 01 × 21 = 21

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

Find the last two digits of 19²⁶⁶.

19²⁶⁶ = (19²)¹³³. Now, 19² ends in 61 (19² = 361) therefore, we need to find the last two digits of (61)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 33²⁸⁸.

33²⁸⁸ = (33⁴)⁷². Now 33⁴ ends in 21 (33⁴ = 33² × 33² = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21⁷². By the previous method, the last two digits of 21⁷² = 41 (tens digit = 2 × 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 87⁴⁷⁴.

87⁴⁷⁴ = 87⁴⁷² × 87² = (87⁴)¹¹⁸ × 87² = (69 × 69)¹¹⁸ × 69 (The last two digits of 87² are 69) = 61¹¹⁸ × 69 = 81 × 69 = 89

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

Find the last two digits of:

1. 27⁴⁵⁶
2. 79⁸³
3. 583⁵¹²

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 24² ends in 76 and 2¹⁰ ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, 24³⁴ will end in 76 and 24⁵³ will end in 24.

Let’s apply this funda:

Find the last two digits of 2⁵⁴³.

2⁵⁴³ = (2¹⁰)⁵⁴ × 2³ = (24)⁵⁴ (24 raised to an even power) × 2³ = 76 × 8 = 08

(NOTE: Here if you need to multiply 76 with 2ⁿ, then you can straightaway write the last two digits of 2ⁿ because when 76 is multiplied with 2ⁿ the last two digits remain the same as the last two digits of 2ⁿ. Therefore, the last two digits of 76 × 2⁷ will be the last two digits of 2⁷ = 28)

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

Find the last two digits of 64²³⁶.

64²³⁶ = (2⁶)²³⁶ = 2¹⁴¹⁶ = (2¹⁰)¹⁴¹ × 2⁶ = 24¹⁴¹ (24 raised to odd power) × 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 62⁵⁸⁶.

62⁵⁸⁶ = (2 × 31)⁵⁸⁶ = 2⁵⁸⁶ × 3⁵⁸⁶ = (2¹⁰)⁵⁸ × 2⁶ × 31⁵⁸⁶ = 76 × 64 × 81 = 84

Find the last two digits of 54³⁸⁰.

54³⁸⁰ = (2 × 3³)³⁸⁰ = 2³⁸⁰ × 3¹¹⁴⁰ = (2¹⁰)³⁸ × (3⁴)²⁸⁵ = 76 × 81²⁸⁵ = 76 × 01 = 76.

Find the last two digits of 56²⁸³.

56²⁸³ = (2³ × 7)²⁸³ = 2⁸⁴⁹ × 7²⁸³ = (2¹⁰)⁸⁴ × 2⁹ × (7⁴)⁷⁰ × 7³ = 76 × 12 × (01)⁷⁰ × 43 = 16

Find the last two digits of 78³⁷⁹.

78³⁷⁹ = (2 × 39)³⁷⁹ = 2³⁷⁹ × 39³⁷⁹ = (2¹⁰)³⁷ × 2⁹ × (39²)¹⁸⁹ × 39 = 24 × 12 × 81 × 39 = 92

Now try to find the last two digits of
1. 34⁵⁷⁶
2. 28²⁸⁷