Every other vertex shares $5$ neighbors with fixed vertex 
An undirected graph has $30$ vertices; one of them is $v$. Each vertex $w\neq v$ shares exactly $5$ neighbors with $v$. Is it true that some vertex must have odd degree?

We could try counting the triples $(v,a,b)$ such that $(v,a),(a,b)$ are edges. Each $b$ yields $5$ vertices $a$, giving $145$ triples, while each $a$ yields $\deg a-1$ triples, giving $\sum_{a\neq v}\deg a-29$ triples, implying
$$\sum_{a\neq v}\deg a=174.$$
But it could still be that $\deg a$ is even for all $a$.
 A: Assume all vertices have even degree.  Then, (a) the set of neighbors, $N$ say, of $v$ has even size, and (b) the vertices outside of $N \cup \{v\}$, call this $M$, has size $29-|N|$, which is odd.
We count the total number of stubs (edge-endpoints) in $N$: if every vertex in $N$ has even degree, the total number of stubs should be even.  We count:


*

*Each $N$-to-$N$ edge contributes $2$ stubs (an even number in total).

*Each $v$-to-$N$ edge contributes $1$ stub (an even number in total).

*Each $M$-to-$N$ edge contributes $1$ stub (an odd number, $5|M|$, in total).


This gives a contradiction.
This is illustrated below:

A: Let $d_i$ be a degree of vertex $v_i$ in $G=\{v, v_1,v_2,...,v_{29}\}$ and let $N(v) = \{v_1,v_2,...v_k\}$ 
Make a bipartite graph $H=(A,B)$ where $B=G\setminus \{v\}$ and $$A=\{\{v,v_1\},...,\{v,v_{29}\}\}$$
We connect a pair $\{v,v_i\}$ with $v_j$ iff in a starting graph $G$ $v$ and $v_i$ are neigbours of $v_j$. So in $H$ each pair has degree $5$ and each vertex $v_j$ has degree $d_j-1$ if $j\leq k$ and $0$ if $j>k$. So we have $$\sum _{i=1}^k (d_i-1)= 29\cdot 5$$ or $$\sum _{i=1}^k d_i= 145+k \;\;\;\;(*)$$


*

*If $k$ is odd, then $v$ has odd degree and we are done.

*If $k$ is even then right side of $(*)$ is odd so if all degrees on the left side are even we get a contradiction. So again some degree is odd again. 

