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

What are the last two digits of 31786?

Solution: 31786 = (30 + 1)786 = 786C × 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  × 1786and 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:

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

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