Perfect numbers!

Let $A$ be the natural number such that $A = 2^{500000000000978}\cdot 4^{929850984395}\cdot 29^{908756489738}$

Then we get that $A = 2^{500000000000978}\cdot (2^2)^{929850984395}\cdot 29^{908756489738}$

$\Rightarrow A = 2^{500000000000978+2\cdot929850984395}\cdot 29^{908756489738}$

$\Rightarrow \boxed {A = 2^{x} \cdot 29^{y}}$ (no need to write the exponents...)

So since $2$ and $29$ are both prime numbers, every single divisor of $A$ is of the form: $2^{n}\cdot 29^{k}$ for $n \in 0, 1, 2, ... (x-1), x$ and $k \in 0, 1, 2, ... (y-1), y$

Thus by using some simple combinatorics we get that $A$ has $(x+1)\cdot (y+1)$ divisors including itself.

Hence all the divisors are: $(2^0\cdot 29^0), (2^0\cdot 29^1), (2^0\cdot 29^2),...(2^x\cdot 29^{y-2}), (2^x\cdot 29^{y-1}), (2^x\cdot 29^y)$

Now, by the definition of the perfect number ("A number is perfect when the sum of its divisors (except the number itself) equals the given number.") we assume that the following equation is true:

$2\cdot A = (2^0\cdot 29^0)+(2^0\cdot 29^1)+(2^0\cdot 29^2)+...+(2^x\cdot 29^{y-2})+(2^x\cdot 29^{y-1})+(2^x\cdot 29^y)$

$\Rightarrow 2\cdot A = 2^0\cdot(29^0+29^1+...+29^{y-1}+29^y)+2^1\cdot(29^0+29^1+...+29^{y-1}+29^y)+...+2^x\cdot(29^0+29^1+...+29^{y-1}+29^y)$

$\Rightarrow 2\cdot A = (29^0+29^1+...+29^{y-1}+29^y)\cdot (2^0+2^1+...+2^{x-1}+2^x)$

So by the sum formula of geometric progression we get:

$2\cdot A = \frac{1\cdot (1 - 29^{y+1})}{1 - 29}\cdot\frac{1\cdot (1 - 2^{x+1})}{1 - 2}$

$\Rightarrow 2\cdot A = \frac{(29^{y+1} - 1)}{28}\cdot (2^{x+1} - 1)$

$\Rightarrow 8\cdot A = \frac{(29^{y+1} - 1)}{7}\cdot (2^{x+1} - 1)$

And it's true that $[29^{y+1}-1]\equiv 0 (mod 7) \Rightarrow [(28+1)^{y+1}-1] \equiv 0 (mod 7)$ and we can prove this through the binomial theorem such that: $(28+1)^{y+1} = \binom{y+1}{0}28^{y+1}\cdot1^0 + \binom{y+1}{1}28^{y}\cdot1^1 + ... + \binom{y+1}{y}28^1\cdot1^y + \binom{y+1}{y+1}28^0\cdot1^{y+1}$

$\Rightarrow (28+1)^{y+1} = \binom{y+1}{0}28^{y+1}\cdot1^0 + \binom{y+1}{1}28^{y}\cdot1^1 + ... + \binom{y+1}{y}28^1\cdot1^y + 1$

And by substituting we get: $[\binom{y+1}{0}28^{y+1}\cdot1^0 + \binom{y+1}{1}28^{y}\cdot1^1 + ... + \binom{y+1}{y}28^1\cdot1^y + 1 -1] \equiv 0 (mod7)$

$28\cdot [\binom{y+1}{0}28^{y}\cdot1^0 + \binom{y+1}{1}28^{y-1}\cdot1^1 + ... + \binom{y+1}{y}28^0\cdot1^y ]\equiv 0 (mod7)$ which is true since $7|28$ and $\frac{(29^{y+1}-1)}{7} = 4\cdot z$ for $z = \binom{y+1}{0}28^{y}\cdot1^0 + \binom{y+1}{1}28^{y-1}\cdot1^1 + ... + \binom{y+1}{y}28^0\cdot1^y$

Thus in the main equation we get:

$8\cdot A = 4\cdot z \cdot (2^{x+1} - 1)$

$\Rightarrow 2\cdot A = z\cdot (2^{x+1} - 1)$

However $2\cdot A$ is an even number while $z\cdot (2^{x+1} - 1)$ is not since both $z$ and $(2^{x+1}-1)$ are odd numbers:

$2^{x+1}\equiv 0 (mod2)\Rightarrow (2^{x+1}-1)\equiv 1 (mod 2)$ and...

$z = \binom{y+1}{0}28^{y}\cdot1^0 + \binom{y+1}{1}28^{y-1}\cdot1^1 + ... + \binom{y+1}{y-1}28^1\cdot 1^{y-1} + \binom{y+1}{y}28^0\cdot1^y$

$z = 28\cdot (\binom{y+1}{0}28^{y-1}\cdot1^0 + \binom{y+1}{1}28^{y-2}\cdot1^1 + ... + \binom{y+1}{y-1}28^0\cdot 1^{y-1}) + \binom{y+1}{y}28^0\cdot1^y$

$z = 28\cdot (\binom{y+1}{0}28^{y-1}\cdot1^0 + \binom{y+1}{1}28^{y-2}\cdot1^1 + ... + \binom{y+1}{y-1}28^0\cdot 1^{y-1}) + (y+1)$ and because $y$ is even $(2\cdot 5 + 8\equiv 8(mod 10)$ <--last digits of the exponents $)$ then $z$ is odd.

Hence, since the above equation is not true then $A$ is not as perfect as it seems!

Observe the number:

$2^{500000000000978}\cdot 4^{929850984395}\cdot 29^{908756489738}$

Obviously, it's an even number. Now, analyze the sum of its proper divisors. Because 29 is an odd prime number, so the given number has:

$908756489738 + 1$ = $908756489739$ odd proper divisors

As we know, sum of an odd number of odd numbers is an odd number. Furthermore, sum of an odd number and an even number is an odd number. Therefore, sum of the given number's proper divisors is an odd number. But the given number is an even number, so it isn't a perfect number.