Why $\mathrm e^{\sqrt{27}\pi } $ is almost an integer? Is there any reason why the expression  $\mathrm e^{\sqrt{27}\pi } $ is almost an integer ??
$ e^{\sqrt{27}\pi }=12288743.98 4 $
Is there an infinite set of numbers with integers 'a' and 'b' so
$ e^{\pi  \sqrt{a}}= b+c $
and 'c' is a real number very close to 1 , for example c=0.99....
Does this method work with other numbers? For example given a (positive) real number d we have that $ d^{\pi \sqrt{a}} $ is an integer for certain values of 'a'
 A: Yes, there is a reason. There are 9 fundamental discriminants $d$ with class number $h(-d)=1$, but there are 4 non-fundamental ones. Hence, the j-function is also an integer for these, namely,
$$\begin{aligned}
j(\sqrt{-3}) &= 2\cdot 30^3\\
j(\sqrt{-4}) &= 66^3\\
j(\sqrt{-7}) &= 255^3\\
j\big(\tfrac{1+\sqrt{-27}}{2}\big) &= -3\cdot 160^3\\
\end{aligned}$$
with the last one explaining your question on $e^{\pi\sqrt{27}}\approx3\cdot 160^3+743.984.$ You can calculate the j-function using this WolframAlpha command.
There are of course other discriminants $d$. For example, the prime-generating polynomial, 
$$P(n) =6n^2+6n+31$$ 
is prime for $n = 0\; \text{to}\; 28$. It has $d = \sqrt{-177}$. And we have,
$$e^{(\pi/3)\sqrt{177}}= 1060^2+9.999929\dots$$
The fact that the “excess” is close to 10 (and not 744) means the integer part does not involve the j-function, but a related function. There are many others.  See also “The j-function and its Cousins”.
A: There is a set of integers $n$ called the Heegner numbers which give almost-integers when you take $e^{\pi\sqrt x}$.  
More specifically, these numbers are square-free $n$ such that the imaginary quadratic field $\mathbb Q[\sqrt{-n}]$ has class number $1$.  It can be shown from this property that $e^{\pi\sqrt{n}}$ becomes very close to an integer for large enough $n$.  
Unfortunately, there are only finitely many Heegner numbers ($1,2, 3, 7, 11, 19, 43, 67, 163$), and only the last few are 'large enough'.  $e^{\pi\sqrt1}$, for example, is $23.14$, hardly an 'almost-integer'.  However, $e^{\pi\sqrt{163}}=262 537 412 640 768 743.99999999999925...$, so the numbers don't have to be that large to give numbers that are extremely close to integers.  
$27$ is not a Heegner number, but you're naturally going to end up getting other numbers $m$ such that $e^{\pi\sqrt m}$ is close to being an integer.  I would guess that the fractional part of $e^{\pi\sqrt m}$ is distributed more or less randomly: then, with probability $1$, and for any $\varepsilon>0$ there will be infinitely many integers $m$ such that $e^{\pi\sqrt m}$ is within $\varepsilon$ of an integer.  
