MHB Calculating M and the Unit Digit of M^2003

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Unit
Click For Summary
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.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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 $$M^{2003}$$.
 
Mathematics news on Phys.org
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:
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 29 ·
Replies
29
Views
3K