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    Prove,using direct arguement, the a four digit number, formed by writing down two digits, and then reversing them, is divisible by 11.


    So,for example, 2112, or 3993.
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    (Original post by Hippysnake)
    Prove,using direct arguement, the a four digit number, formed by writing down two digits, and then reversing them, is divisible by 11.


    So,for example, 2112, or 3993.
    Let x and a represent these numbers then the number is axxa = a(10^3)+x(10^2)+x(10^1)+a = a(10^3+1)+x(10+100) jaja you get the idea
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    One more- For any pair of numbers, x and y, 2(x^2 + y^2) is the sum of two squarea.
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    (Original post by Hippysnake)
    One more- For any pair of numbers, x and y, 2(x^2 + y^2) is the sum of two squarea.
    Clearly 2(x^2+y^2) = 2x^2 + 2y^2 = (x*sqrt(2))^2 + (y*sqrt(2))^2 ... fairly obvious from there I think
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    (Original post by Necro Defain)
    Clearly 2(x^2+y^2) = 2x^2 + 2y^2 = (x*sqrt(2))^2 + (y*sqrt(2))^2 ... fairly obvious from there I think
    Is that not it? That's the answer right?

    EDIT: How can 2x^2 be written as (Root2 x X^2)? It doesn't work.
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    the answer is 5.
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    (Original post by Hippysnake)
    Is that not it? That's the answer right?

    EDIT: How can 2x^2 be written as (Root2 x X^2)? It doesn't work.
    It's (root2 x X)^2, kinda hard to write that out but Idk how to use latex well
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    (Original post by Necro Defain)
    It's (root2 x X)^2, kinda hard to write that out but Idk how to use latex well
    If you expand that though do you not get something completelt different? And neg to the other guy for being a ****.
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    (Original post by Hippysnake)
    If you expand that though do you not get something completelt different? And neg to the other guy for being a ****.
    Nope... (root(2)*x)^2 = (root(2))^2 * (x)^2 = 2x^2
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    Any number is divisible by 11 if the 1st, 3rd, 5th digits add up to the sum of the 2nd, 4th, 6th digits etc, or the difference is a multiple of 11.

    Ie 15323 is divisible by 11 because 1+3+3 = 5+2

    Obviously the rule can be extended to any size integer.
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    (Original post by Hippysnake)
    One more- For any pair of numbers, x and y, 2(x^2 + y^2) is the sum of two squarea.
    So for 1 and 1, 4 is the sum of two squares?


    I think they have to be different numbers.

    I usually stick in a pair of numbers to see if I can see a pattern.

    3 and 2 gives us 26, which is 5^2 + 1^2. There is a pattern there.
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    (Original post by Necro Defain)
    Clearly 2(x^2+y^2) = 2x^2 + 2y^2 = (x*sqrt(2))^2 + (y*sqrt(2))^2 ... fairly obvious from there I think
    A square is usually a square of an integer...

    OP:
    I wish I could be more help, but I can't without giving it away. Just experiment writing 2(x2+y2) in different ways.
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    A bit of algebra:
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    2x^2 + 2y^2 = (x+y)^2 + (x-y)^2
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    (Original post by meatball893)
    A square is usually a square of an integer...

    OP:
    I wish I could be more help, but I can't without giving it away. Just experiment writing 2(x2+y2) in different ways.
    Don't have a clue....
 
 
 
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