We all know how to find the last digit using cyclicity when we are given a number raised to a power. Last digit of a number depends only on the last digit of the base. You must be quite familiar with something like this –

Last Digit of Base:

0 – Last digit of expression with any power will be 0.

1 – Last digit of expression with any power will be 1.

2 – 2, 4, 8, 6, 2, 4, 8, 6… Cyclicity is 4.

3 – 3, 9, 7, 1, 3, 9, 7, 1… Cyclicity is 4.

4 – 4, 6, 4, 6, 4, 6, 4, 6… Cyclicity is 2.

5 – Last digit of expression with any power will be 5.

6 – Last digit of expression with any power will be 6.

7 – 7, 9, 3, 1, 7, 9, 3, 1… Cyclicity is 4.

8 – 8, 4, 2, 6, 8, 4, 2, 6… Cyclicity is 4.

9 – 9, 1, 9, 1, 9, 1, 9, 1… Cyclicity is 2.

Cyclicity is nothing but pattern recognition. You see that when you multiply 2 by itself, there is a pattern of last digit which goes 2, 4, 8, 6, 2, 4, 8, 6 and so on. We can use the same principle for when a question asks us for the last two digits of the expression. Let me remind you first that here at QWQW, we sometimes flirt with the lines that define GMAT scope. Obviously, we do point out whenever we are indulging and that’s exactly what we are going to do in this post. We are carrying on for the love of Math and the Q51 score.

The last two digits of the base decide the last two digits of the expression. For example,

**Example 1: **Let’s look at powers of 11.

11^1 = 11

11^2 = 121

11^3 = 1331

11^4 = …41 (we should just multiply the last two digits together and ignore the rest)

11^5 = …51

11^6 = …61

11^7 = …71

Note that the last two digits are displaying a pattern depending on the power. So we expect the cyclicity here to be 10.

11^8 = …81

11^9 = …91

11^(10) = …01

11^(11) = …11

11^(12) = …21

and so on. So the last two digits should go from 11, 21 to 91, 01 and then go to 11 again. The cycle of 10 starts from power of 1, 11, 21 etc. This means that 11^(46) should have last two digits as 61, 11^(92) should have last two digits as 21 and 11^(168) should have last two digits as 81.

Let’s look at some other numbers now:

**Example 2:** Say, we need the last two digits of 6^{58}

6^1 = 6 (No second last digit)

6^2 = 36

6^3 = 216

6^4 = …96 (Just multiply the last two digits)

6^5 = …76

6^6 = …56

6^7 = …36

and hence starts the cycle again:

3, 1, 9, 7, 5, 3, 1, 9, 7, 5 and so on.

The new cycle with tens digit of 3 begins at the powers of 2, 7, 12, 17, 22, 27 etc. So the new cycle will also begin at power of 57 and 6^58 will have 1 as the tens digit.

**Example 3:** How about the last two digits of 7^102?

7^1 = 7 (No second last digit)

7^2 = 49

7^3 = 343

7^4 = …01

7^5 = …07

7^6 = …49

7^7 = …43

We see a cyclicity of 4 here: 49, 43, 01, 07, 49, 43, 01, 07 … and so on. The new cycle begins at 2, 6, 10, 14 i.e. even powers which are not multiples of 4. So a new cycle will begin at 102 too. So the last two digits of 7^(102) will be 49.

Now there can be many variations in the questions asking us to find the last two digits. We will use different concepts for different question types. Today we saw how to use pattern recognition. We will look at some other methods next week.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*