In this article we will discuss about the concept of last two digits of a number. We know that to get the unit digit of a number, we do not need the whole number but only the last digit of the number. Similarly, to get the last two digits we do not the whole number but the last two digits of the number are required. For example, to get the last two digits of the product 3124 × 5628, we just need the product of last two digits of these numbers i.e., 24 × 28 = 672 or 72 will be the last two digits of 3124 × 5628.

Now finding the last two digits of the numbers of the form a^{n} is not that simple as finding the last digit of such numbers. In order to understand this concept better you have to learn some other basic results.

First of all, let us check the squares of some numbers:

21^{2} = 441, 22^{2} = 484, 23^{2} = 529, 24^{2} = 576, 25^{2} = 625, 26^{2} = 676, 27^{2} = 729, 28^{2} = 784, 29^{2} = 841

You can check that after 25, the last two digits of the squares of the numbers are repeating, i.e., the last two digits of 26^{2} are same as that of 24^{2}, similarly, last two digits of 27^{2} are same as that of 23^{2} and so on.

In fact, the general rule is that, the last two digits of the square of a number say ‘a’ is same as the last two digits of the square of the number (50 – a), (50 + a), (100 – a), (100 + a) and so on.

Therefore, the last two digits of 24^{2} will be same as last two digits of (50 – 24)^{2} i.e., 26^{2} or (50 + 24)^{2} i.e., 74^{2} or (100 – 24)^{2} i.e., 76^{2} and so on. Now, this is true for the last two digit of every number and it is very important as we are going to use it in almost every question on finding the last two digits.

Next, you must remember the following results:

**(i) When the number is of the form (a1) ^{ xy…..z}. **

The last two digits of the number of the form (a1)^{ xy…..z} = a×z1.

It means that the unit digit of such numbers will always be 1 but the digit at the ten’s place will be the unit digit of the product obtained by multiplying ‘a’ with the unit digit of the power.

Note that, here xy……z is any number and we need the unit digit of this number in the power.

E.g., the unit digit of the number (81)^{123} is 1 and the ten’s place digit is given by 8 × 3 = 24 or 4.

Therefore, the last two digits of (81)^{123} = 41.

Let us take another example:

The last two digits of the number (21)^{321548} = 61, where 6 is the unit digit of the product of 2 (of 21) and 8 (the unit digit of the power).

**(ii) When number is of the form (a5) ^{ m} where m is a natural number greater than 1.**

The last two digits of the number of the form (a5)^{ m} = 25, if at least one among ‘a’ and ‘m’ is even

The last two digits of the number of the form (a5)^{ m} = 75, if both ‘a’ and ‘m’ are odd.

E.g., the last two digits of (25)^{125} = 25 as here 2 (of 25) is even. but the last two digits of (35)^{125} = 75 as here both 3 (of 35) and the power are odd.

**(iii) When number is of the form (24) ^{ m}.**

The last two digits of the number of the form (24)^{m} = 24, if m is odd

The last two digits of the number of the form (24)^{m} = 76, if m is even.

Also, 76^{m} where m is any natural number always ends at 76.

While solving the questions of last two digits, you must keep the above mentioned rules in mind and also the concept of the last two digits of the squares of the numbers mentioned above.

**Ex. Find the last two digits of 2 ^{100}.**

**Sol: **Whenever the question asks about the last two digits of the numbers of the form 2^{m}, always break 2^{m} as (2^{10})^{n} as 2^{10} is 1024 or ends at 24.

So, we can write 2^{100} = (2^{10})^{10} = (24)^{10} = 76 (As 24^{even number} ends in 76 and also only the last two digits of 1024 are taken as we do not need the complete number)

Therefore, the last two digits of 2^{100} are 76.

**Ex.** **Find the last two digits of 3 ^{244}.**

**Sol:** Whenever the question asks about the last two digits of the numbers of the form 3^{m}, always break 3^{m} as (3^{4})^{n} as 3^{4} is 81.

Therefore, 3^{244} = (3^{4})^{61} = (81)^{61} = 81 (Check the rule of (a1) ^{xy…….z} above)

**Ex.** **Find the last two digits of (17) ^{256}.**

**Sol:** In these type of questions, try to convert the number in the form (a1)^{xy….z} by taking the squares of the numbers repeatedly and using the result for the last two digits of the squares of the numbers.

We have (17)^{256} = (17^{2})^{128} = (89)^{128} (Taking last two digits of 289)

= (89^{2})^{64} = (21)^{64} (Square of 89 will end where square of 100 – 89 = 11 will end)

= 81 (Check the rule of (a1) ^{xy…….z} above)

Now that you have understood the concept and the ways to apply the above results. Let us practice the following questions.

**1. Find the last two digits of (33) ^{146}.**

**Sol:** Here (33)^{146} = (33^{2})^{73} = (89)^{73} (Square of 33 will end where square of 50 – 33 = 17 will end)

= (89^{2})^{36} × 89 = (21)^{36} × 89 = 21 × 89 = 1869.

Therefore, the last two digits of (33)^{146} are 69.

**2. Find the last two digits of (59) ^{250}.**

**Sol:** We have (59)^{250} = (59^{2})^{125} = (81)^{125} (Square of 59 will end where square of 50 + 9 or 9 will end)

= 01 (Digit at the ten’s place is the unit digit of the product 8 × 5 = 40).

**3. Find the last two digits of 48 ^{118}.**

**Sol:** Here 48^{118} = (2^{4} × 3)^{118} = 2^{472} × 3^{118} …………(i)

Now 2^{472} = (2^{10})^{47} × 2^{2} = (24)^{47} × 4 = 24 × 4 = 96 (Odd power of 24 ends at 24)

Again, 3^{118} = (3^{4})^{29} × 3^{2} = (81)^{29} × 9 = 21 × 9 = 89 (Only last two digits taken)

Therefore, from (i) 48^{118} = 2^{472} × 3^{118} = 96 × 89 = 8544 or 44.

**4. Find the last two digits of the number 72 ^{255}.**

**Sol:** Here 72^{255} = (2^{3} × 3^{2})^{255} = 2^{765} × 3^{510} …………..(i)

Now 2^{765} = (2^{10})^{76} × 2^{5} = 24^{76} × 32 = 76 × 32 = 32 (Only last two digits taken)

Again, 3^{510} = (3^{4})^{127} × 3^{2} = (81)^{127} × 9 = 61 × 9 = 49 (Only last two digits taken)

Therefore, from (i), 72^{255} = (2^{3} × 3^{2})^{255} = 2^{765} × 3^{510} = 32 × 49 = 1568 or 68.

**5. Find the last two digits of the number 53 ^{454} × 64^{258}.**

**Sol:** We have 53^{454} × 64^{258} = 53^{454} × (2^{6})^{258} = 53^{454} × 2^{1548} …………(i)

Now 53^{454} = (53^{2})^{227} = (09)^{227} = (9^{2})^{113} × 9 = (81)^{113} × 9 = 41 × 9 = 69

Also 2^{1548} = (2^{10})^{154} × 2^{8} = 24^{154} × 56 (only last two digits of 2^{8} are taken)

= 76 × 56 = 56 (Only last two digits taken)

Therefore, 53^{454} × 64^{258} = 69 × 56 = 3864 or the last two digits are 64.

Hope the above discussion is helpful to you to understand the concept of last two digits.

In the next article we will discuss another concept of number system for CAT examination.

Till then Happy learning!!!

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