# Program to find the last two digits of x^y

**The task is to find the last two digits of x^y.**

Since the digits with which it can end are 0-9, Hence **this problem can be divided into 5 cases:**

**Case 1: when x ends with 1**For finding the last two digit of a number, when the number ends with 1 then we have to do following steps shown as in the figure.

**Example: 21^44**So, Last two digit of 21^44 is 81.

So, Last two digit of 31^35 is 51.

**Case 2: when x ends with 3, 7, 9**For finding the last two digit of a number, when the number ends with 3, 7, 9 then we have to apply cyclicity concept to convert the last digit as a 1.

**cyclicity of 3:**3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1**cyclicity of 7:**7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

**cyclicity of 9:**9^1 = 9

9^2 = 1**Example1: 23^34**

**Solution:**- Last digit of 23^34 is 3 so, we use cyclicity of 3 .
- 3^4 gives 1 so, we take 23^4
- ((23)^4)^8 * (23)^2
- last two digit of (23)^4) is 41, so we take (41)^8 and solve according to the given diagram.
- So last digit of (41)^8 is 21 .
- solve (23)^2, the last digit of (23)^2 is 29.
- Now multiply last digit of (41)^8 i.e 21 with the last digit of (23)^2 i.e 29
- i.e 21 * 29 = 609
- So, Last two non zero digit of 23^34 is 09.

**Example2: 37^45**

**Solution:**- Last digit of 37^45 is 7 so, we use cyclicity of 7 .
- 7^4 gives 1 so, we take 37^4
- ((37)^4)^11 * (37)^1
- last two digit of (37)^4) is 61, so we take (61)^11 and solve according to the diagram.
- So last digit of (61)^11 is 61 .
- solve (37)^1, the last digit of (37)^1 is 37.
- Now multiply last digit of (61)^11 i.e 61 with the last digit of (37)^1 i.e 37
- i.e 61 * 37 = 2257
- So, Last two non zero digit of 37^45 is 57.

**Example3: 59^22**

**Solution:****Case 3: when x ends with 2, 4, 6, 8**For finding the last digit of a number ends with 2, 4, 6, 8; We use number 76 which is a type of magic number because its square, cube and etc contain last 2 digit numbers as itself i.e 76.

**Take an example**:

square of 76 = 5776, its last two digit =76

cube of 76 = 438976, its last two digit=76So we take two cases:

- if (2^10)^even power then it always return 76 .
- if (2^10)^odd power then it always return 24 .

**Steps for finding last two digits**- Firstly, convert given number in to these formats if (2^10)^power. Here power will be odd or even according to the question.
- Now, check power will be odd or even.
- if power is odd then its value will be 24.
- if power is even then its value will be 76.

**Examples****Example1:**Find last 2 digit of 2^453.**Solution:****Step 1:- conversion**

2^453 = (2^10)^45 * 2^3**Step 2:- odd power so we take 24**

= 24 * 8

= 192

So, Last two non zero digits of 2^453 are 92.

**Example2:**Find last 2 digits of 4^972.**Solution:****step 1:- conversion**

4^972 = (2^2)^972

= 2^1944

= (2^10)^194 * 2^4**step 2:- even power so, we take 76**

= 76 * 16

= 1216

So, Last two non zero digits of 4^972 are 16.

**Example3:**Find last 2 digits of 6^600.**Solution:****step 1:-**conversion

6^600 = (2)^600 * (3)^600

= (2^10)^60 * ((3)^4)^150 {Apply case 2 in (3)^600}**step 2:-**(2^10)^60 has even power so, we take 76 as the last digit**step 3:-**Solve ((3)^4)^150, we get 01 as the last digit**step 4:-**last digit of (2^10)^60 i.e 76 multiply with the last digit of ((3)^4)^150 i.e 01**step 5:-**i.e 76 * 01 = 76

So, Last two non zero of 6^600 is 76.

**Example4:**Find last 2 digits of 8^330.**Solution:****step 1:- conversion**

8^33 = (2^3)^110

= (2)^330**step 2:-**(2^10)^33 has odd power so, we take 24 as the last digit

So, Last two non zero digits of 8^330 are 24.

**Case 4: when x ends with 5**For finding the last two digit of a number, when the number ends with 5 then we have to follow the table which is given below.

**Example1:**Find last 2 digit of 25^25.**Solution:**- first digit of number is 2 i.e even
- Last digit of a power is 5 i.e odd
- Now, even-odd combination gives last digit as a 25

So, the last two non zero digits of 25^25 are 25.

**Example2:**Find last 2 digit of 25^222.**Solution:**- first digit of number is 2 i.e even
- Last digit of a power is 2 i.e even
- Now, even-even combination gives last digit as a 25

So, the last two non zero digits of 25^222 are 25.

**Example3:**Find last 2 digit of 165^222.**Solution:**- first digit of number is 1 i.e odd
- Last digit of a power is 2 i.e even
- Now, odd-even combination gives last digit as a 25

So, the last two non zero digits of 165^222 are 25.

**Example4:**Find last 2 digit of 165^221.**Solution:**- first digit of number is 1 i.e odd
- Last digit of a power is 1 i.e odd
- Now, odd-odd combination gives last digit as a 75

So, the last two non zero digits of 165^221 are 75.

**Case 5: when x ends with 0**For finding the last two digit of a number, when the number ends with 0 then we have to check next digit and according to the digit calculate the last digit.

**Example:**Find last 2 digit of 150^221.**Solution:**- 150 last digit is 0 so we check next digit i.e 5 and apply case 4
- first digit of number is 1 i.e odd
- Last digit of a power is 1 i.e odd
- Now, odd-odd combination gives last digit as a 75

So, the last two non zero digits of 165^221 is 75.

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