Question 1204795
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I'm assuming that b=(a-1)^1/2 +1 should be the equation b=(a-1)^(1/2) +1, which is the same as writing {{{b=(a-1)^(1/2)^"" +1}}} and converts to {{{b = sqrt(a-1)+1}}}


List of the first few primes = {2,3,5,7,11,13,17,19,23,29,31,...}
There are infinitely many prime numbers.


Let's say we picked the smallest prime for value "e".
Doing so will guarantee that a+b+c+d+e is minimized.


e = (d-1)^(1/2) + 1
(d-1)^(1/2) = e-1
d-1 = (e-1)^2
d = (e-1)^2 + 1
d = (2-1)^2 + 1
d = 2


e = 2 leads to d = 2
But d must be different from e, so we have to go back to the drawing board.


Let's pick the next largest prime.
e = 3
d = (e-1)^2 + 1
d = (3-1)^2 + 1
d = 5
We get a different value this time. So far so good.
Furthermore, the result is prime.


Then,
d = (c-1)^(1/2) + 1
c = (d-1)^2 + 1
c = (5-1)^2 + 1
c = 17
We get a different prime.


And,
c = (b-1)^(1/2) + 1
b = (c-1)^2 + 1
b = (17-1)^2 + 1
b = 257
which is also prime. 
Primality can be checked using software. Or you can check the primes {2,3,5,7,11,13} to find they aren't factors of 257.


And lastly,
b = (a-1)^(1/2) + 1
a = (b-1)^2 + 1
a = (257-1)^2 + 1
a = 65537
This value is prime.
Use software to check primality.


To summarize
a = 65537
b = 257
c = 17
d = 5
e = 3
all of which are different primes and fit the equations mentioned.
This is the smallest set possible since e = 3 was chosen among the smallest primes.


So,
a+b+c+d+e = 65537+257+17+5+3 = <font color=red size=4>65819</font> is the smallest sum possible, which leads to <font color=red size=4>answer choice D</font>
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