# (ABC)^2=CCCDE

genius
Posers and Puzzles 13 Oct '02 12:22
1. genius
Wayward Soul
13 Oct '02 12:22
(ABC)^2=CCCDE each letter stands for a different didgit, and C is
twice E. what does each letter stand for???
2. 13 Oct '02 13:28
a=2
b=9
c=8
d=0
e=4

right?
3. genius
Wayward Soul
13 Oct '02 16:02
that is correct, and a challenge to other people-why??? sintubin has
told us the answer, now you must tell us why!!!
4. Jacko
Knock, Knock...?
13 Oct '02 21:28
you got ur homework done alan is that not enough

David
5. genius
Wayward Soul
14 Oct '02 08:56
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...
6. Jacko
Knock, Knock...?
14 Oct '02 10:15
i got no email
7. Schliemann
The Diplomat
25 Oct '02 09:06
Hmm..a double negative?

Dave
(Just poking fun)
8. legionnaire
Free Thinker
14 Oct '02 19:58
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
9. 14 Oct '02 20:06
in this set of 10 (88804, 88814, ....) there can only be one number
that has an integer root. that is 88804.