*Thursday, October 12th, 2017*

Nike caused controversy with its advertising campaign during the 1996 Olympics by using the slogan, “You Don’t Win Silver — You Lose Gold.” Nike’s use of this slogan drew harsh criticism from many former Olympic Silver medallists. In a way, it did undermine the importance of the second position but in Math things are often very different. Figuring out the second last digit is often tougher than figuring out the last digit. It is unlikely but definitely not impossible that in CAT you get a straightforward question that asks you to find out the second last digit of a number (abc^{pqr}). It did happen in CAT 2008. In few cases, you will be able to do it by forming a cycle and observing the pattern. Those will be the easier cases. Read on if you wish to do the same for the not so easy cases.

The question becomes really simple if the last digit in abc^{pqr} is 0 or 5 because if it 0, second last digit will be 0 and if it is 5, second last digit will be 2 or 7 (which can be easily figured out by observing the cyclicity). All the other questions can be divided in two broad categories:

a) Last digit is odd

b) Last digit is even

*I recommend that before using any of the concepts given below, you should try and see if a pattern exists. *

Let us consider our number is abc^{pqr} where a,b,c,p,q and r are digits and c is not 0 or 5.

** ****Concept 1: What to do when the last digit is odd?**

**The second last digit always depends on the last two digits of the number so anything before that can be easily neglected. **

**We first convert the number in such a way that the last digit of the base becomes 1. Second last digit of the number will simply be:**

*Last digit of (Second last digit of base) X (Last digit of power)*

**Let us look at few examples**

**Eg 1a:** Second last digit of 3791^{768} = Last digit of 9×8 = 2

**Eg 1b:** Second last digit of 1739^{768} = Second last digit of 39^{768} = Second last digit of Second Last digit of 1521^{384 }= Last digit of 2 × 4 = 8

**Eg 1c:** Second last digit of 9317^{768} = Second last digit of 17^{768} = Second last digit of (17^{4})^{ 192} = Second last digit of (…21)^{ 192} = Last digit of 2 x 2 = 4

** ****Concept 2: What to do when the last digit is even?**

**The second last digit always depends on the last two digits of the number so anything before that can be easily neglected. **

We need to remember the following ideas:

- **2 raised to power 10 will always end in 24.
- 24 raised to an even power will always end in 76 and to an odd power will always end in 24.
- 76 raised to any power will always end in 76.

Now we can use these to find out the second last digit. We reduce the number in such a way that the last two digits of the base become 76.

**Eg 2a:** Second last digit of 1372^{482}

- Second last digit of 72
^{482} - Second last digit of 72
^{480}x 72^{2} - Second last digit of (72
^{10})^{ 48}x (**84) - Second last digit of 24
^{48}x (**84) - Second last digit of 76 x 84
- Second last digit of 6384 = 8

**Eg 2b:**** **Second last digit of 48^{307 }_{=} (48^{3})^{ 102} x 48 = (****92)^{ 102} x 48

- Second last digit of 92
^{100}x 92^{2}x 48 = 76 x (**64) x 48 - Second last digit of (****72) = 7

**Eg 2c:**** **Second last digit of 154^{84} = Second last digit of (54)^{ 84}

- Second last digit of (54
^{5})^{ 16}x 54^{4}= (***24)^{ 16 }x (54^{2})^{ 2} - Second last digit of 76 x (2916)
^{ 2} - Second last digit of 76 x 56
- Second last digit of 4256 = 5

I hope that after reading this post you will be at ease in figuring out the second last digit. I also hope that you will not mind winning silver medals either.