# Exponents and end digits of powers

*Power* is the product of multiplying a number by itself. We say 16 is the power of 2 to the 4th power. This can be written as 2^{4}= 16. The number 2 is called the *base* and 4, the number of times the base number is taken as factor is called the *exponent*. Here is a problem about powers and exponents.

###### Problem

Study the following number patterns and note the last digit of the powers.

4^{2} = 4 x 4 = 16

4^{3} = 4 x 4 x 4 = 64

4^{4} = 4 x 4 x 4 x 4 = 256

The last digit of 4^{2} is 6, that of 4^{3} is 4 and that of 4^{4} is 6.

1. Which of the following will also be equal to a number with last digit 4? Show/Give your reason.

A. 4^{16} B. 4^{18} C. 4^{21 } D. 4^{40}

2. What is the last digit of the power of 4^{2011}?

3. What is the last digit of 2^{2012}? Show your solution.

###### Answers

You can tell from the exponents in the questions that you are not expected to do multiplication. No ordinary calculator can help you in these questions. So please don’t even attempt. Questions like these call for noticing patterns.

If you cannot make a conjecture based on the given cases in the problem you can always generate other cases. For example you can find the last digit of 4^{5}and then 4^{6 }which is 4 and 6 respectively. Hmmm … so they alternate. If your answer is C and reason that you just alternate 4 and 6 until you eliminated choices A and B, I’ll give you a 2 out of 4 marks even if you are correct. But if you say your answer is C because you observed that the last digit is 4 when the exponent is odd and the last digit is 6 when the exponent is even then you will get the full mark.

Question 2 is similar to question 1. And question 3? Well, you can find the pattern in the last digit of powers of 2 (Warning: it’s tedious) or you can simply solve it like this:

The exponent of 4 is even so the last digit must be 6.

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