Let $\color{#69047E}{Cody}$ start from $(0,0)$ .

$\bullet$ Then, his displacement along $\color{#D61F06}{\textbf{X axis}}$ is actually

$\color{#3D99F6}{+100 - 1 + \dfrac{1}{100} - \dfrac{1}{10^4} ... \infty}$

$\bullet$ and displacement along $\color{#D61F06}{\textbf{Y axis}}$ is actually

$\color{#20A900}{+10 - \dfrac{1}{10} + \dfrac{1}{10^3}-\dfrac{1}{10^5} ... \infty}$

These are 2 GPs , $\color{#3D99F6}{a_1=100} , \color{#20A900}{a_2=10}$ and $\color{#3D99F6}{r_1=} \color{#20A900}{r_2 =} \color{#D61F06}{\dfrac{-1}{100}}$

By using sum of infinite terms of a GP ,i.e. $sum_\infty=\dfrac{a}{1-r}$ , we get that

$\color{#3D99F6}{S_x = \dfrac{100}{1-(-\frac{1}{100})} = \dfrac{10000}{101}}$

$\color{#20A900}{S_y = \dfrac{10}{1-(-\frac{1}{100})} = \dfrac{1000}{101}}$

Actual displacement = $\sqrt{\color{#3D99F6}{S_x ^2} +\color{#20A900}{ S_y ^2}} = \sqrt{\dfrac{10^8+10^6}{101^2}}$

Hence the displacement is $\sqrt{\dfrac{10^6(10^2+1)}{101^2}}=\sqrt{\dfrac{10^6}{101}} = \dfrac{1000}{\sqrt{101}}$

Hence asked thing is $101\times 1000 = \boxed{101000}$ .

The question, though seemingly twisty, is actually easy. Notice that cody moves with distance of $100,10,1,0.1.....$ which form a G.P. Now, our aim is to find the distance. lets consider the horizontal displacement first. We notice that its $100-1+0.01-0.0001......$ uptill infinity. Now, we can surely find its sum by geometric progression formula, as $\frac{100}{1+0.01}$ ...

Now, lets try to find out vertical displacement. Notice that its also $10-0.1+0.001......$ , which again is a GP with a sum of $\frac{10}{1+0.01}$

Thus, we can the actual displacement is $\sqrt{(\frac{100}{1+0.01})^2+(\frac{10}{1+0.01})^2}=\frac{1000}{\sqrt{101}}$

Thus, the answer is $101000$

Also, if Cody moves generally in the same way with distances as $a,ar,ar^2,ar^3,......$ , Then the displacement after infinite turns will be
$\frac{a}{\sqrt{1+r^2}}$

As in other solutions, Cody begins his jogging at (x,y)=(0,0). His position converges rapidly to around (x,y)=(99.0099009901, 9.90099009901). We hazard the guess that this is x=10000/101 and y=1000/101, hence the solution.

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def change ( ) :
    possibilities = [    
        lambda x, y, c : ( x + c, y ), # E
        lambda x, y, c : ( x, y + c ), # N
        lambda x, y, c : ( x - c, y ), # W
        lambda x, y, c : ( x, y - c ), # S
        ]
    state = 0
    yield possibilities [ state ]
    while True :
        state = ( state + 1 ) % 4
        yield possibilities [ state ]

def run ( ) :
    state = 100
    yield 100
    while True :
        state *= 0.1
        yield state

c = change ( )
r = run ( )
x, y = 0, 0
for i in range ( 15 ) :
    x, y = c . next ( ) ( x, y, r . next ( ) )
    print x, y

break into x-component and y-component as you would do it in physics.

I use Wolfram and sum geometrical progression

$1+k+k^2+... = 1/(1-k), k=1/10^4$

With aid of a complex plane,
Final position (with starting point as origin) $=100+10i-1-\cdots=\dfrac{100}{1-0.1i}$
Magnitude= $\dfrac{100}{\sqrt{1^2+0.1^2}}=\dfrac{1000}{\sqrt{101}}$
$a\times b=1000\times 101=101000$