MHB Calculating M and the Unit Digit of M^2003

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The discussion revolves around calculating the value of M defined as M = (3x - 1)/(1 + x) - (√|x| - 2 + √(2 - |x|))/(|2 - x|). It is determined that x must be -2, leading to M being calculated as 7. The unit digit of M^2003 is found by evaluating M^3 mod 10, resulting in a unit digit of 3. The calculations involve modular arithmetic, specifically noting that M^4 ≡ 1 mod 10. The conversation also includes appreciation for problem-solving and contributions to mathematical discussions.
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Let $$x$$ be a real number and let $$M=\frac{3x-1}{1+x}-\frac{\sqrt{\mid x\mid-2}+\sqrt{2-\mid x \mid}}{\mid 2-x \mid}$$.

Find $$M$$ and also the unit digit of $$M^{2003}$$.
 
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anemone said:
Let $$x$$ be a real number and let $$M=\frac{3x-1}{1+x}-\frac{\sqrt{\mid x\mid-2}+\sqrt{2-\mid x \mid}}{\mid 2-x \mid}$$.

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

-Dan
 
anemone said:
Let $$x$$ be a real number and let $$M=\frac{3x-1}{1+x}-\frac{\sqrt{\mid x\mid-2}+\sqrt{2-\mid x \mid}}{\mid 2-x \mid}$$.

Find $$M$$ and also the unit digit of $$M^{2003}$$.

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
 
kaliprasad said:
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! :o:p

topsquark said:
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 $$\cap$$ humorous member at MHB!:cool:
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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