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zmudz

  • one year ago

Compute the number of ordered pairs of integers \((x,y)\) with \(1\le x<y\le 100\) such that \(i^x+i^y\) is a real number.

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  1. anonymous
    • one year ago
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    \[x=\left\{ 2m,m \in N,1\le m \le 49 \right\}\] \[y=\left\{ 2n,n \in N,2\le n \le 50 \right\}\] \[\left( x,y \right)=\left\{( 2m,2n),m<n,m \in N,n \in N,1\le m \le 49,2\le n \le 50 \right\}\]

  2. thomas5267
    • one year ago
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    But x=1 and y=3 would also work right?

  3. zmudz
    • one year ago
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    @surjithayer @thomas5267 , I got (1, 3), (3, 1), (0, 2), and (2, 0) but 4 ordered pairs is wrong :( I can't get the answer unless I give up on the problem, but I really need the points. I'm not sure what @surjithayer 's notation in the set means. Any further help is greatly appreciated!

  4. thomas5267
    • one year ago
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    \(0\leq x\color{red}{<}y\leq100\)

  5. thomas5267
    • one year ago
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    \(i\) is the imaginary number right?

  6. anonymous
    • one year ago
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    x and y are even numbers when m=1,x=2*1=2 when n=2 y=2*2=4 \[\iota ^2=-1\]

  7. thomas5267
    • one year ago
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    Nope, x and y being even is not the only cases when \(i^x+i^y\) is real. \(-i+i=0\in\mathbb{R},\,1+1=2\in\mathbb{R}\)

  8. thomas5267
    • one year ago
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    \[ \begin{align*} x&=2n,\,1\leq n\leq49\\ y&=2m,\,2\leq n+1\leq m \leq 50\\ n&=1,\,x=2,\,y=4,6,8,\dots,100\\ n&=1,\,x=2,\,49\text{ choices of }y\\ n&=2,\,x=4,\,y=6,8,\dots,100\\ n&=2,\,x=4,\,48\text{ choices of }y \end{align*} \] Choices of \((x,y)\) with \(x=2n,\,1\leq n\leq49,\,y=2m,\,2\leq n+1\leq m \leq 50\): \(49+48+47+\cdots+3+2+1=50\times49\div 2=1225\) \[ \begin{align*} x&=2n-1,\,1\leq n\leq49\\ y&=2m-1,\,2\leq n+1\leq m \leq 50\\ n&=1,\,x=1,\,y=3,5,7,\dots,99\\ n&=1,\,x=1,\,49\text{ choices of }y\\ n&=2,\,x=3,\,y=5,7,9,\dots,99\\ n&=2,\,x=3,\,48\text{ choices of }y \end{align*} \] Choices of \((x,y)\) with \(x=2n-1,\,1\leq n\leq49,\,y=2m-1,\,2\leq n+1\leq m \leq 50\): \(49+48+47+\cdots+3+2+1=50\times49\div 2=1225\) \[ \text{Total} = 1225\times2=2450 \]

  9. thomas5267
    • one year ago
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    Something is wrong with my answer.

  10. thomas5267
    • one year ago
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    \[ \begin{align*} x&=4n-3,\,1\leq n\leq 25\\ y&=4m-1,\,1\leq n\leq m \leq 25\\ n&=1,\,x=1,\,y=3,7,11,\dots,99\\ n&=1,\,x=1,\,25\text{ choices of }y\\ n&=2,\,x=5,\,y=7,11,15,\dots,99\\ n&=2,\,x=3,\,24\text{ choices of }y \end{align*} \] Choices of \((x,y)\) with \(x=4n-3,\,1\leq n\leq 25,\,y=4m-1,\,1\leq n\leq m \leq 25\): \(25+24+23+22+\cdots+2+1=26\times25\div2= 325\) \[ \begin{align*} x&=4n-1,\,1\leq n\leq 24\\ y&=4m-3,\,2\leq n+1\leq m \leq 25\\ n&=1,\,x=3,\,y=5,9,13,\dots,97\\ n&=1,\,x=3,\,24\text{ choices of }y\\ n&=2,\,x=7,\,y=9,13,17,\dots,97\\ n&=2,\,x=3,\,23\text{ choices of }y \end{align*} \] Choices of \((x,y)\) with \(x=4n-1,\,1\leq n\leq 24,\,y=4m-3,\,2\leq n+1\leq m \leq 25\): \(24+23+22+21+\cdots+2+1=25\times24\div2= 300\) Total=\(300+325+1225=1850\) This is the answer returned by Mathematica.

  11. zmudz
    • one year ago
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    1850 is the right answer. The teacher got it by doing casework: (a) \(i^x = i\) and \(i^y = -i\), or (b) \(i^x = -i\) and \(i^y = i\) They both give 625 cases, because there are 25 numbers n for which \(i^n = i\) (namely, \(n = 1, 4, \ldots, 97\)), and there are 25 numbers n for which \(i^n = -i\) (namely \(n = 3, 7, \ldots, 99\)). This gives 1250 cases total. There are 2500 other cases (50*50) that result from having any pair of even numbers in which it will give a real answer. But then there are overlaps of 50 cases when x=y, so you subtract that. And by symmetry, only half of the remaining will satisfy x<y, so that gives 1850.

  12. thomas5267
    • one year ago
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    Basically the same idea but your teacher count all combinations then subtract duplicates where I count all possible combinations directly at a cost of a higher complexity.

  13. zmudz
    • one year ago
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    yep

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