LAST TWO DIGIT

Dear student, Agenda is how to calculate last two digits of the number ending on 1.
Last digit or unit digit will be always 1.
Just observe tenth place digit in following example
11^2 = 121 (1 * 2 = 2)
11^3= 1331 (1 * 3 = 3)
21^2 = 441 (2 * 2 = 4)
21^3= 9261 (2 * 3 = 6)
Just apply this technique to the following examples
Example: 31^8
If number is ending on 1 last digit is always 1.Important is how to calculate Tenth place digit. It is also very easy. Tenth place digit will be (3 * 8 =24) so tenth place digit is 4
So last two digits of 31^8 is 41.
Find the last two digits of 41^2789
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 71^56747
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 51^456 × 61^567
The last two digits of 51^456 will be 01 and the last two digits of 61^567 will be 21. Therefore, the last two digits of 514^56 × 61^567 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.
19^266 = (19^2)133. Now, 19^2 ends in 61 (19^2 = 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 33^288.
33^288 = (33^4)72. Now 33^4 ends in 21 (33^4 = 33^2 × 33^2 = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21^72. By the previous method, the last two digits of 21^72 = 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:
Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method.
Now try the method with a number ending in 7:
Find the last two digits of 87^474.
87^474 = 87^472 × 87^2 = (87^4)^118 × 87^2 = (69 × 69)^118 × 69 (The last two digits of 87^2 are 69) = 61^118 × 69 = 81 × 69 = 89
If you understood the method then try your hands on these questions:
Find the last two digits of:

1. 27^496

2. 79^93

3. 583^512