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Something interesting concerning parity... 
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#1
Sep1810, 03:02 PM

P: 154

Is there a proof or way of proving that all even numbers (taking into account the definition of an even number as [tex]n=2k[/tex]) end in 0,2,4,6, or 8?



#2
Sep1810, 07:49 PM

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P: 11,155

What do you think? Care to try showing that numbers ending in 1,3,5,7, or 9 can not be even?



#3
Sep1910, 05:05 AM

P: 24

Rewrite n as 10a+b, where 0<=b<10.
Then n is even means 2 divides 10a+b, i.e. 2 divides b (let 10a+b=2k and solve for b). Hence b=0,2,4,6,8 


#4
Sep1910, 10:09 AM

P: 154

Something interesting concerning parity...
Is there a way to show that the fact that all even numbers end in 0,2,4,6, or 8 implies the definition of an even number ([tex]n=2k[/tex], where [tex]k[/tex] is an integer)? 


#5
Sep1910, 06:50 PM

P: 185

Any number that is a multiple of ten is even and divisible by 2. Then any number (base ten) denoted by ...dcba (where a is the ones place, b in the tens, etc.) will be divisible by two if a is divisible by 2 because ...+d*10^3+c*10^2+b*10^1+a represents the number and division is linear. All terms except a are multiples of ten always and therefore divisible by 2, then all that is left is a.



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