1. Standard membergenius
    Wayward Soul
    Your Blackened Sky
    Joined
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    15128
    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. Joined
    01 Dec '01
    Moves
    14745
    13 Oct '02 13:28
    a=2
    b=9
    c=8
    d=0
    e=4

    right?
  3. Standard membergenius
    Wayward Soul
    Your Blackened Sky
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    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. DonationJacko
    Knock, Knock...?
    St Andrews, Scotland
    Joined
    18 Mar '02
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    38882
    13 Oct '02 21:28
    you got ur homework done alan is that not enough

    David
  5. Standard membergenius
    Wayward Soul
    Your Blackened Sky
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    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. DonationJacko
    Knock, Knock...?
    St Andrews, Scotland
    Joined
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    38882
    14 Oct '02 10:15
    i got no email
  7. Standard memberSchliemann
    The Diplomat
    Slightly Left :D
    Joined
    22 Jun '01
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    8518
    25 Oct '02 09:06
    Hmm..a double negative?

    Dave
    (Just poking fun)
  8. Donationlegionnaire
    Free Thinker
    New York City
    Joined
    22 Mar '02
    Moves
    10815
    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. Joined
    01 Dec '01
    Moves
    14745
    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.
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