(ABC)^2=CCCDE each letter stands for a different didgit, and C is
twice E. what does each letter stand for???

a=2
b=9
c=8
d=0
e=4

right?

that is correct, and a challenge to other people-why??? sintubin has
told us the answer, now you must tell us why!!!

you got ur homework done alan is that not enough

David

but i figured it out before i posted it here-and to prove it, i will give a
clue.
5*5=25
35*35= somethingaruther, but it will end in a 5...
7*7=49
37*37=somethingaruther, but it will end in a 7...
that is the clue. anyway-i e-mailed you the answer, jacko...

i got no email

Hmm..a double negative?

Dave
(Just poking fun)

Well, since C has to be twice E, the only single digits that are twice
another single digit are 2, 4, 6 and 8. No matter how many digits
there are in a number, if you square it, the last number is the last
digit of the square of the last digit of the number you're squaring.
(for example, the last digit of 559 squared must be 1, since 9x9 is
81). So the only number out of those four whose last digit of its
square is half the number itself is 8. Therefore, C must be 8, and E
must be 4.

After that I crapped out, you're left with possibilities: 88804, 88814,
etc., and I just plugged them into a calculator. Let me know if there's
a more elegant method for figuring out the other digits.

-legionnaire

in this set of 10 (88804, 88814, ....) there can only be one number
that has an integer root. that is 88804.