show that $\displaystyle 11^{33}-11^{31} $ is exactly divisible by 5

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- Aug 28th 2010, 07:49 PMPunchshow that 11^33-11^31 is divisible by 5
show that $\displaystyle 11^{33}-11^{31} $ is exactly divisible by 5

- Aug 28th 2010, 07:56 PMundefined
- Aug 28th 2010, 07:57 PMPunch
LOL i didnt understand a thing....

how about using simple methods, methods that are below or equal to o levels requirement - Aug 28th 2010, 08:09 PMundefined
- Aug 28th 2010, 08:11 PMPunch
i still dont understand... why does 11 raised to any exponent n>0 will end in a 1?

- Aug 28th 2010, 08:14 PMundefined
- Aug 28th 2010, 08:20 PMIsomorphism
If you are looking for an alternative way (I do not know what counts as an o level):

Use the exponent laws and distributive law to rewrite:

$\displaystyle 11^{33} - 11^{31} = 11^{31}11^{2} - 11^{31} = \cdots$

Can you continue? - Aug 28th 2010, 08:24 PMPunch
$\displaystyle 11^{31}[11^2-11]$

- Aug 28th 2010, 08:25 PMPunch
- Aug 28th 2010, 08:27 PMundefined
- Aug 28th 2010, 08:29 PMundefined
You mean power of 11, not multiple of 11. Remember ordinary elementary school subtraction.. start at the units digit... so if the top number has a 1 and so does the bottom number, the difference must have 0 as units digit..... so it's a multiple of 10, as you say even and divisible by 5.