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PageDown

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    Song Lyrics

    in Music

    Posted

    Anyone can help me remember the artist who sang the song "I'll Always Love You?"

     

    The lyrics goes like this:

     

    I'll always love you

    Yes, I will

    Through all the good times

    Throught all the things we've cried

    Baby, I almost died

    Trying to find you

    But hen you came through

    And open my eyes....

     

    Sing it to me softly

    sing it to me slow

    La la la la

    La la la la...

    Oh, I always love you

    All of my life through

    This is my song...

     

    Sorry kung mejo mali-mali ang lyrics... he he he

    Math Problems!........share It Here!

    in Academics, Schools, Universities

    Posted

    246

     

    I don't know enough of number theory, so this is what I used instead to get the answer (it's real clunky, so apologies :) ):

     

    The largest sum possible for two 3-digit numbers is 1998, so _IF_ the sum of the mystery number and 596 results in a 4 digit number then this 4 digits would look like either of these two (since they are in geo. progression):

     

        b        b*r      b*r^2    b*r^3

    b*r^3    b*r^2      b*r        b

     

    (where each digit <=9)

     

    With the leftmost digit (either 'b' or 'b*r^3') being a 1 (since we're checking the case where the sum is a 4 digit number).

     

    So now we have either of these two conditions (since the leftmost digit is a 1):

     

    b=1

    b*r^3=1

     

    In the first case, if b=1 then r=1 or 2  (since in this case, the rightmost digit 'b*r^3' must be <=9, thus r can at most only be equal to 2). Thus the possible 4-digit numbers here would be 1111 (b=1,r=1) and 1248 (b=1,r=2). Subtracting 596 from these yields 515 and 652, niether of which are in arith. progress. So this case is false.

     

    If we take the second case, b*r^3=1, then b=r=1. Which again yields the possible 4-digit number 1111, subtracting 596 gives 515 which again isn't in arith. progress. so this case is false as well.

     

    So the above shows that the sum of the mystery number and 596 cannot be a 4-digit number.

     

    Ok, so now we know the sum is a 3-digit number, hence in the following form:

     

        b      b*r      b*r^2

    b*r^2    b*r      b

     

    We also know each digit <=9 and since 596 is being added the leftmost digit must >=6, thus:

     

    9 >=    b    >= 6

    9 >= b*r^2 >= 6

     

    Taking the second case, b*r^2, there are only 3 possible values allowed for r (due to the <=9 condition for each digit), hence drawing a matrix for values of b given the allowed values for r:

     

    b*r^2    b (r=1)    b (r=2)    b (r=3)

    -------    --------    ---------    ---------

      6            6              X              X

      7            7              X              X

      8            8              2              X

      9            9              X              1

     

    ('X' marks non-valid combinations)

     

    Which gives 6 possible 3-digit numbers for the sum of the mystery number and 596:

     

    666 (b=6,r=1)

    777 (b=7,r=1)

    888 (b=8,r=1)

    999 (b=9,r=1)

    842 (b=2,r=2)

    931 (b=1,r=3)

     

    Subtracting 596 from each will show that only 842 will give you a 3-digit number (i.e. 246) which has it's digits in arith. progress.

     

    Now for the first case, 9>=b>=6 ... nah, I'm too lazy, besides I've already got at least one answer :)

     

    Nope! you have to use a 4 x 4 matrix to get to it (easier solution).

    Math Problems!........share It Here!

    in Academics, Schools, Universities

    Posted

    Sorry, you've got the correct answer but a BAD solution, as if you domn't know anything about math.

     

    The answer is easy, but it's the way you solved it.

     

    A don't see any progression formula.

     

    When you say a 3-digit number, than it should be written as:

    Let x = 100's digit, y = 10s didigt, and z = 1s digit

    So the number would be 100x+10y+z = the number

     

    This is how you solve a math problem... Ok?

    The Motel Amenities Thread - merged threads.

    in Motels and Apartelles Topics

    Posted

    They're right! It's not all fiction, it's purely f#&k este FACT pala.

     

    It's in VC Pasay, along Cuneta Ave.. My companion and I went on a f#&king date kaya lang isinama pa pati ang friend. Since wala akong magawa, I tagged them along, 2 girls and me.

     

    The room has 2 floors, the bedroom which has a round bed and upstairs, the jacuzzi with TV.

     

    My date and I stayed in the ground floor para lumindol man, d ramdam ng friend nya na nasa itaas..

     

    Nice room.

     

    One time, my girlfriend went to VC Cuneta and had a room with 2 floors, again (?). Downstairs was the dining room and shower with jacuzzi while upstairs was the main bedroom.

     

    But the best place i've been to was in VC Pasay (way back). It has a spacious and luscious area. Why? The room is divided by a bridge (huh, bridge in a room). The 1st part was the main bedroom with kitchen while the other part had a sauna (really, a sauna) and jacuzzi, a large toilet and bathroom (with love seat at that!). Galeng noh!

     

    I dunno if it's still there....

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