# show that 11^33-11^31 is divisible by 5

• Aug 28th 2010, 07:49 PM
Punch
show 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 PM
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Quote:

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

Easy with modular arithmetic

$\displaystyle 11^{33}-11^{31} \equiv 1^{33}-1^{31} \equiv 1 - 1 \equiv 0 \pmod{5}$
• Aug 28th 2010, 07:57 PM
Punch
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 PM
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Quote:

Originally Posted by Punch
LOL i didnt understand a thing....

how about using simple methods, methods that are below or equal to o levels requirement

All right how's this: We know that 11 raised to any exponent n > 0 will end in a 1. So 11^n - 11^k for n,k>0 must end in a 0, hence be divisible by both 2 and 5.
• Aug 28th 2010, 08:11 PM
Punch
i still dont understand... why does 11 raised to any exponent n>0 will end in a 1?
• Aug 28th 2010, 08:14 PM
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Quote:

Originally Posted by Punch
i still dont understand... why does 11 raised to any exponent n>0 will end in a 1?

Consider the way you do long multiplication on paper. The units digit of 11 is a 1, so when you multiply it by 11 you get units digit 1, and when you multiply that number (121) bu 11 you again get units digit 1, etc. Try it and see.
• Aug 28th 2010, 08:20 PM
Isomorphism
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 PM
Punch
$\displaystyle 11^{31}[11^2-11]$
• Aug 28th 2010, 08:25 PM
Punch
Quote:

Originally Posted by undefined
Consider the way you do long multiplication on paper. The units digit of 11 is a 1, so when you multiply it by 11 you get units digit 1, and when you multiply that number (121) bu 11 you again get units digit 1, etc. Try it and see.

so u mean that each time i multiply by 11, i get an 1 on the end of the number and when i minus another multiple of 11, i am deducting that 1 and so the number becomes even and is divisible by 5
• Aug 28th 2010, 08:27 PM
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Quote:

Originally Posted by Punch
$\displaystyle 11^{31}[11^2-11]$

Should be $\displaystyle 11^{31}[11^2-1]$. Now evaluating $\displaystyle 11^2 - 1$ will tell you something...
• Aug 28th 2010, 08:29 PM
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Quote:

Originally Posted by Punch
so u mean that each time i multiply by 11, i get an 1 on the end of the number and when i minus another multiple of 11, i am deducting that 1 and so the number becomes even and is divisible by 5

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.