Hi All,

I am not quite sure if such questions appear on the actual GMAT test, but I remember struggling with one such question during my practice test earlier. Came across this logic on web and thought of sharing. Please share if there is something easier than this to crack such questions.
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Logic for deriving last digits of a particular complicated numerical can be divided into 3 major categories -

1) When the last digit in abc raised to pqr is 0 or 5
2) Last digit is odd
3) Last digit is even

Category 1 is fairly simple -
because if it 0, the 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).

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 raised to pqr where a, b, c, p, q and r are digits and c is not 0 or 5.

Category 2: 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. The second last digit of the number will then 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 raised to 768 = Last digit of 9×8 = 2
Eg 1b: Second last digit of 1739 raised to 768 = Second last digit of 39 raised to 768 = Second last digit of Second Last digit of 1521 raised to 384 = Last digit of 2 × 4 = 8
Eg 1c: Second last digit of 9317 raised to 768 = Second last digit of 17 raised to 768 = Second last digit of (17 raised to 4) raised to 192 = Second last digit of (…21) raised to 192 = Last digit of 2 x 2 = 4

Category 3: 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 raised to 482
→ Second last digit of 72 raised to 482
→ Second last digit of 72 raised to 480 x 72 raised to 2
→ Second last digit of (72 raised to 10) raised to 48 x (**84)
→ Second last digit of 24 raised to 48 x (**84)
→ Second last digit of 76 x 84
→ Second last digit of 6384 = 8

Eg 2b: Second last digit of 48 raised to 307 = (48 raised to 3) raised to 102 x 48 = (****92) raised to 102 x 48
→ Second last digit of 92 raised to 100 x 92 raised to 2 x 48 = 76 x (**64) x 48
→ Second last digit of (****72) = 7

Eg 2c: Second last digit of 154 raised to 84 = Second last digit of (54) raised to 84
→ Second last digit of (545) raised to 16 x 54 raised to 4 = (***24) raised to 16 x (54 raised to 2) raised to 2
→ Second last digit of 76 x (2916) raised to 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 in such type of questions.

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Keep studying hard!
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Rgds,
Mudit
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'Take risks and conquer your fears!'