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So the question is:

is there a graph G on 50 vertices that is a tree and has a total degree of 100?

What I got so far was:

number of vertices = edges + 1

$50 = x + 1$

$ x = 49 $

$\frac{49}{100} = 0.49$

Therefore, no because 49 isn't divisible by 100.

Am I correct or am I wrong? Please correct me if I'm wrong.

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    You are correct. – 2017-02-17
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    What is the "total degree" of a graph?? – 2017-02-20

2 Answers 2

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By this, you have $$\sum_{v\in V}deg(v) = 2|E| = 100,$$ then $|E| = 50$, but $$|E| \leq |V| - 1 = 49.$$

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This answer assumes the "total degree" of the graph is equal to the sum of the vertex degrees.

As you state, a tree on $50$ vertices has $49$ edges. But we needn't get into "indivisibility" afterwards, we can simply continue...

Since each edge has two endpoints, the total degree is $2 \times 49=98 \neq 100$.

The first claim has come up a number of times on math.SE, e.g.:

Proof verification: Prove that a tree with n vertices has n-1 edges
Proving that A connected graph on $n$ vertices is a tree iff it has $n-1$ edges
Prove by induction the predicate (All n, n >= 1, any tree with n vertices has (n-1) edges).
Proof for "A simple connected graph has n-1 edges iff it is a tree " without induction.
Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.

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    Your answer seems to be based on the assumption that by "total degree": of a graph the OP means the sum of the degrees of the vertices. You may be right, but it seemed unlikely to me for two reasons. First, the OP seems to think that the number of edges is divisible by the "total degree". Second, if he meant the sum of the degrees, why wouldn't he have just *said* the sum or the degrees (or the degree-sum)? – 2017-02-20
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    You're right. I made this assumption explicit now; it's hard to tell what "total degree" means, but what else could it mean besides "sum of vertex degrees"? – 2017-02-20