# Last Two Digits of a Number

The crux of the article is to help understand the concept and solving questions like, find the last two digits of

1. $${346}^{4}$$
2. $${ 375 }^{ 49 }$$
3. $${ 31}^{123}$$

### To easily understand this topic, we will break it into four categories:

1. Last two digit of numbers ending with one.
2. Last two digit of numbers ending with 3,7,9.
3. Last two digit of numbers ending with 5.
4. Last two digit of numbers ending with 2,4,6,8.

## CASE 1:Let us first discuss the numbers ending with one.

Let us take an example to have a clear understanding of the concept

Find last two digits of $$51^{36}$$

As to everyone’s knowledge, it is clear that one raised to any power will give you 1 only, hence it is clear that the last digit of the answer will be 1.

Now we need to find the tens digit of the answer. To find the tens digit of you need to multiply the tens digit of the base to the units digit of exponent 5×6=30.

Now, take units digit of this product i.e. 0. 0 will be the final tens digit.

Last two digit of $$51^{36}$$ is 01.

## CASE 2:Now, comes the numbers ending with 3,7,9.

NOTE: In order to understand this case, we need to know last two digits of $${ (100-a) }^{ 2 }$$

Now lets understand CASE 2 with an example.

Solve for last two digits of $${ 279 }^{ 24 }$$

The questions ending with 3,7,9 can easily be converted into questions with numbers ending in one.

In the question, we need to find $${ 279 }^{ 24 }$$, therefore let us take the last two digits of the number 279 i.e., 79 and write the exponent i.e., 24 as (2×12)

Using the concept given in image above, we can say:

$$({ 79 }^{ 2 })^{ 12 }$$ = $$[(100-21)^{ 2 }]^{ 12 }$$ = $${ (21 }^{ 2 })^{ 12 }$$ = $$(41)^{ 12 }$$ = 81 (using CASE 1)

Last two digit of $${ 279 }^{ 24 }$$ = 81

## CASE 3:Last two digits of a number ending with 5.

For numbers ending with 5, if the tens digit of the base and the ones digit of the exponent are both odd then the last two digit of the number will be 75. For all other cases, the last two digits will be 25.

### Following are some examples for last digits of numbers ending in 5:

1. $${ (575) }^{ 87 }=75$$
2. $${ 165 }^{ 43 }=25$$
3. $${ 685 }^{ 123 }=25$$
4. $${ 235 }^{ 59 }=75$$

## CASE 4 :Last two digits of numbers ending with 2,4,6,8.

### Following points will make this case easy for you:

1. There is only one even two-digit number which always ends in itself – 76 i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as the last two digits for even numbers. We know that $${ 24 }^{ 2 }$$ ends in 76 and $${ 2 }^{ 10 }$$ 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, $${ 24 }^{ 34 }$$  will end in 76 and $${ 24 }^{ 53 }$$ will end in 24.
2. $${ ({ 2 }^{ 10 }) }^{ odd }$$ ends in 24 and $${ ({ 2 }^{ 10 }) }^{ even }$$ends in 76
3. $${ 2 }^{ n }$$ multiplied by 76 gives last two digits as $${ 2 }^{ n }$$

### Let us apply this concept in an example to understand the case.

#### Q) Find last two digits of $${ (264) }^{ 236 }$$

Taking all the 2’s out from $${ (264) }^{ 236 }$$

$$({ 2 }^{ 3 }\times 33)^{ 236 }$$ = $$({ 2 }^{ 708 }\times 33^{ 236 })$$

Now, this question can be solved in two parts:

### PART 1

$${ 2 }^{ 708 }$$ = $${ (2 }^{ 10 })^{ 70 }\times { 2 }^{ 8 }$$

As we know that $${ (2 }^{ 10 })^{ even }$$ gives us 76.

Therefore the solution can be further continued as follow:

76$${ \times 2 }^{ 8 }$$ = 76$$\times 56$$

(NOTE: $${ 2 }^{ n }$$ multiplied by 76 gives last two digits as $${ 2 }^{ n }$$.

Hence part 1 is solved and last two digits obtained for $${ 2 }^{ 708 }$$ are 56.

### PART 2

To solve $${ 33 }^{ 236 }$$ we need to use the theory of numbers ending in 3,7 and 9. $${ (50-a) }^{ 2 }$$ ends in same last two digits as $${ a }^{ 2 }$$ .

$${ 33 }^{ 236 }$$ can be written as $${ (50-17) }^{ 236 }$$

It could be further written as :

$$[{ (50-17) }^{ 4 }]^{ 59 }$$

=> $$[[{ (50-17) }^{ 2 }]^{ 2 }]^{ 59 }$$

=> $$[[{ (17) }^{ 2 }]^{ 2 }]^{ 59 }$$

=> $$[{ (89) }^{ 2 }]^{ 59 }$$

which can again be simplified in the format  $${ (100-a) }^{ 2 }$$

$$[{ (100-11) }^{ 2 }]^{ 59 }$$

=> $$[{ (11) }^{ 2 }]^{ 59 }$$

=> $${ (21) }^{ 59 }$$

By using the method discussed in CASE 1 (numbers ending in 1) we get last two digits here as:

81

Part 2 is solved and the value we got here is 81.

$${ (264) }^{ 236 }$$ = $$({ 2 }^{ 3 }\times 33)^{ 236 }$$ = $$({ 2 }^{ 708 }\times 33^{ 236 })$$ => 56$$\times$$81 => 4536

Now multiplying part 1 with part 2, we get last two digits as 36.

Different cases i.e., numbers with the last two digits 1,3,7 and 9 were covered in this question.

1. Find the units digit of $${ 361 }^{ 123 }$$
2. Find the units digit of $${ 294 }^{ 196 }$$
3. Find the units digit of $${ 355 }^{ 156 }$$

Answers: 1) 81, 2) 56, 3) 25

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