From the notes on $K$-theory from Allen Hatcher, one notices that $K$-theory was used to prove that the only division algebras over $\mathbb{R}$ are the real, complex, quaternion and octonion algebras with division.
Which other strong results from $K$-theory exist that are understandable for an undergraduate student to motivate him/her to study $K$-theory?
The Hairy Ball Theorem states that every vector field on $S^2$ has a zero, unlike the circle $S^1$ which has a nowhere-zero vector field, usually written $\frac{\partial}{\partial\theta}$. So a natural question to ask is which spheres $S^n$ have a nowhere-zero vector field? Given the observations so far, one might guess that it depends on the parity of $n$ (i.e. whether $n$ is even or odd). This turns out to be the case. It follows from the Poincaré-Hopf Theorem that a closed manifold has a nowhere-zero vector field if and only if it has Euler characteristic zero. As $\chi(S^n) = 1 + (-1)^n$, we see that $S^n$ has a nowhere-zero vector field if and only if $n$ is odd.
Now that we know which spheres admit a nowhere-zero vector field, we can try to determine how many such vector fields they admit. Of course, we could just take the nowhere-zero vector field we have and multiply by a non-zero number to obtain another nowhere-zero vector field. To avoid this kind of redundancy, we instead ask a more refined question:
A collection of vector fields $V_1, \dots, V_k$ on a smooth manifold $M$ are said to be linearly independent if $\{V_1(x), \dots, V_k(x)\}$ is linearly independent for every $x$.
If $n$ is even, the answer is zero. If $n$ is odd, the answer is at least $1$, but is at most $n$. If the answer is $n$, then $S^n$ is said to be parallelisable (i.e. the tangent bundle is trivial). The only values of $n$ for which $S^n$ is parallelisable are $n = 0, 1, 3,$ and $7$ which is related to the fact that they are the unit spheres in $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, and $\mathbb{O}$ respectively. So for every other odd value of $n$, we know the answer is at least one but at most $n - 1$.
The highlighted problem was resolved by Frank Adams in his $1962$ paper Vector Fields on Spheres. It had been previously shown that there are at least $\rho(n+1) - 1$ linearly independent vector fields on $S^n$, and Adams showed that it it does not admit $\rho(n+1)$ linearly independent vector fields. Here $\rho(n)$ is the $n^{\text{th}}$ Radon-Hurwitz number: write $n = 2^{4a+b}c$ with $a, b, c$ non-negative integers, $c$ odd, and $0 \leq b \leq 3$, then $\rho(n) = 2^b + 8a$. In his paper, Adams combines homotopy theory and topological $K$-theory in order to solve the aforementioned problem (which lies in the realm of differential topology).
I have listed some values below to give a sense of structure of these seemingly strange Radon-Hurwitz numbers.
$$ \begin{array}{c|ccccccccccccccccc } n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 & 17 & 19 & 21 & 23 & 25 & 27 & 29 & 31 & 33\\ \hline \rho(n) & 1 & 3 & 1 & 7 & 1 & 3 & 1 & 8 & 1 & 3 & 1 & 7 & 1 & 3 & 1 & 9 & 1 \end{array} $$
Here is the first $10,000$ terms, just note that it is indexed differently: the left hand column is the value of $n$ and the right hand column is the value of $\rho(2n + 1)$. One can show that every integer $k \equiv 0, 1, 3, 7 \bmod 8$ is of the form $\rho(n)$ for some $n$. However, for a given $k$ the smallest $n$ for which $\rho(n) = k$ might be much larger than $k$. For example, in the first $10,000$ terms, the highest value of $\rho(n)$ is $27$.