How do you take the multiplicative inverse of a p-adic number? I am reading the wiki page for p-adic numbers and it states that they are a field extension of the rationals so each member has to have a modular multiplicative inverse. 
So how would I take the inverse of, say, 35 in the ring of 2-adic numbers?
 A: Another way: note that $2^{12} \equiv 1 \text{ mod }35$.  Now $(2^{12} - 1)/35 = 117 =  1 + 2^2 + 2^4+2^5+2^6$
so
$$ \dfrac{1}{35} = -\frac{117}{1-2^{12}} = - \sum_{j=0}^\infty\sum_{k \in \{0,2,4,5,6\}} 2^{12 j+k} = 1 + \sum_{j=0}^\infty\sum_{k \in \{1,3,7,8,9,10,11\}} 2^{12j+k}$$
A: Long division.  $1$ divided by $35$.  First, $35$ base $2$ is $100011.$  So the problem is:  

Look at the right-most digits.  $1$ goes into $1$ how many times?  $1$:  

Multiply:  

Subtract:  

Next digit is $1$, so $1$ goes in the quotient:  

Multiply:  

Subtract:  

Next digit is zero, $0$ goes in the quotient.  Next digit is $1$.  Multiply:

Subtract:

Three zeros, then a 1:  

Continue.  It is eventually periodic, of course.
A: Following @RobertIsrael, I say: $35=1+34$. Using the geometric series for $1/(1+x)$,
$$
1/35=1-34+34^2-34^3+\cdots\,,
$$
a $2$-adically convergent series.
A: With Mathematica, the following command will provide you an answer for 20 digits :
BaseForm[PowerMod[35, -1, 2^20], 2]
= 10001010111110001011 (base 2)
