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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
Let \(\displaystyle x\) be a real number and let \(\displaystyle M=\frac{3x-1}{1+x}-\frac{\sqrt{\mid x\mid-2}+\sqrt{2-\mid x \mid}}{\mid 2-x \mid}\).

Find \(\displaystyle M\) and also the unit digit of \(\displaystyle M^{2003}\).
 

topsquark

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MHB Math Helper
Aug 30, 2012
1,123
Let \(\displaystyle x\) be a real number and let \(\displaystyle M=\frac{3x-1}{1+x}-\frac{\sqrt{\mid x\mid-2}+\sqrt{2-\mid x \mid}}{\mid 2-x \mid}\).

Find \(\displaystyle M\) and also the unit digit of \(\displaystyle M^{2003}\).
Just a moment and I'll have it. I just have to program Excel...

-Dan
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Let \(\displaystyle x\) be a real number and let \(\displaystyle M=\frac{3x-1}{1+x}-\frac{\sqrt{\mid x\mid-2}+\sqrt{2-\mid x \mid}}{\mid 2-x \mid}\).

Find \(\displaystyle M\) and also the unit digit of \(\displaystyle M^{2003}\).
We are taking square root of |x| - 2 and its –ve so it has to be zero
So |x| - 2 = 0 or x = 2 or – 2
x cannot be 2 as |2-x| is in denominator
so x = - 2
hence putting the value x = -2 we get M = 7
as M^4 = 1 mod 10
M^2000 = 1 mod 10 or M^2003 = M^3 mod 10 = 343 mod 10 or 3
3 is the unit digit
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
We are taking square root of |x| - 2 and its –ve so it has to be zero
So |x| - 2 = 0 or x = 2 or – 2
x cannot be 2 as |2-x| is in denominator
so x = - 2
hence putting the value x = -2 we get M = 7
as M^4 = 1 mod 10
M^2000 = 1 mod 10 or M^2003 = M^3 mod 10 = 343 mod 10 or 3
3 is the unit digit
Hi kaliprasad,

Thanks for taking the time to participate in this challenge problem and I can tell how much you enjoyed working with some of the problems that I posted here and in case if you have any interesting mathematics problems to share with us, please feel free to do so! :eek::p

Just a moment and I'll have it. I just have to program Excel...

-Dan
Hi Dan,

Thank you for the reply and you know what, you're one of the clever \(\displaystyle \cap\) humorous member at MHB!:cool: