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    find all squares ending in 444.

    this is what i've done so far:

    for a number's square to end in 4 it must have a units digit of 2 (2*2=4), or 8 (8*8=64). first take 2:
    if a number squared (ending with 2) makes a number ending in 44, then twice the product of the tens digit and the ones digit must give a 4. call this digit x:
    2*x*2 = 4 => x=1. this clearly works for 12*12 = 144.
    now to find the hundreds digit.
    the hundreds digit of the square of the number (which we set to 4) depends on twice the product of the hundreds digit and the ones digit (call the hundreds digit = y), PLUS the product of the tens digit. so:
    1*1 + 2*y*4 = 4
    8y = 3
    y = ?
    from this do I conclude that there are no numbers ending in 2 whose square ends in 444?

    I got stopped even quicker when working with the 8*8=64 because I know you carry the 4:
    the tens digit must be x such that:
    6 + x*8*2 = 4
    6 + 16x = 4
    16x = -2 ?


    as you can see I kind of crashed and burned.
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    There are systematic ways of going about solving this type of question with modular arithmetic.

    We'll begin by trying to find all those squares that end in 44.

    This means we're solving

    x^2 =44 (mod 100).

    The Chinese Remainder Theorem tells us that (as 100 = 4 x 25) the problem is the same as solving

    x^2 = 44 (mod 4) = 0 (mod 4)

    and

    x^2 = 44 (mod 25) = 19 (mod 25).

    Now x^2 = 4 (mod 4) means x = 0 or 2 (mod 4) means x = 0 (mod 2).

    Looking at the squares of 0,1,2,...,12 (mod 25) we get

    0, 1, 4, 9, 16, 0, 11, 24, 14, 6, 0, 21, 19.

    So that the only solutions of x^2 = 19 (mod 25) are 12 and -12 (=13).

    We see then that any x whose square ends in 44 is even and is 12 or 13 (mod 25).

    Numbers, between 0 and 99 that are 12 or 13 mod 100 are:

    12, 13, 37, 38, 62, 63, 87, 88

    and of these, the even ones are

    12, 38, 62, 88.

    What this means is that if x^2 ends in 44 then x ends in one of the four above.

    So a number, whose square ends 444 is itself of the form

    n12, n38, n62, n88

    Looking at the first

    (100n + 12)^2 = 10000n^2 + 2400n + 144 = 400n + 144 (mod 1000)

    So 400n + 144 = 444 (mod 1000)

    means 400n = 300 (mod 1000)

    means 4n = 3 (mod 10)

    which has no solutions as 4n will be even and so not end in 3.

    The second case gives

    (100n + 38)^2 = 10000n^2 + 7600n + 1444 = 600n + 444 (mod 1000)

    So 600n + 444 = 444 (mod 1000)

    means 600n = 0 (mod 1000)

    means 6n = 0 (mod 10)

    means n = 0 or 5 (mod 10)

    So any number ending 038 or 538 will have a square ending 444.

    The third case gives

    (100n + 62)^2 = 10000n^2 + 12400n + 3844 = 400n + 844 (mod 1000)

    So 400n + 844 = 444 (mod 1000)

    means 400n = -400 = 600 (mod 1000)

    means 4n = 6 (mod 10)

    means n = 4 or 9 (mod 10).

    So any number ending 462 or 962 will have a square ending 444.

    The fourth case gives

    (100n + 88)^2 = 10000n^2 + 17600n + 7744 = 600n + 744 (mod 1000)

    So 600n + 744 = 444 (mod 1000)

    means 600n = -300 = 700 (mod 1000)

    means 6n = 7 (mod 10)

    which has no solutions as an even number 6n can't end in 7.

    So in summary x must end 038, 462, 538, 962.
 
 
 
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Updated: February 5, 2005

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