The number $\sqrt{53+20\sqrt{7}+148}$ can be written in the form $a+b\sqrt{c}$, where $a,$ $b,$ and $c$ are integers and $c$ has no factors which is a perfect square of any positive integer other than 1. Find $a+b+c$.
We can find $(a+b+c)$ by separating the terms under the square root and simiplyfing to obtain $$53+20\sqrt{7}+148 = (9+4\sqrt{7})(9-4\sqrt{7})$$We know that when a quadratic has two integer roots $\sqrt{b^2-4ac}$ becomes expressible as a real number. We start by finding the roots, completing the square: $$81-112+28=(9+4\sqrt{7})^2-(2\sqrt{112})$$Unfortunately, $2\sqrt{112}$ isn't a real number, so we must find the factors of $112$. $$112=2^4\cdot7$$ Simplifying to $2\sqrt{2^2}\cdot2\sqrt{2}\cdot\sqrt{7}$, we find that $a+b+c=2+2+\boxed{7}= \boxed{11}$.
Let's start by simplifying the expression $\sqrt{53+20\sqrt{7}+148}$.
We notice that $53+148=201$, which can be factored as $9 \times 9 \times 2 + 2 \times 9 \times 2 \times \sqrt{7} + (\sqrt{7})^2$.
This can be further rewritten as $(9+2\sqrt{7})^2$.
So, the given expression $\sqrt{53+20\sqrt{7}+148}$ is equal to $9+2\sqrt{7}$.
Therefore, $a=9$, $b=2$, and $c=7$.
The sum $a+b+c$ is $9+2+7 = \boxed{18}$.