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    Prove that if a number leaves a remainder 2 when it is divided by 3 then its square leaves a remainder 1 when divided by 3
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    (Original post by Zain786H)
    Prove that if a number leaves a remainder 2 when it is divided by 3 then its square leaves a remainder 1 when divided by 3
    Of course if you do this exercise using modular arithmetic it is very simple. If not then if a number x leaves a remainder of 2 when divided by 3 then it is of the form x=3y+2 where y is an integer.
    So x^2=(3y+2)^2=9y^2+12y+4 and you should be able to show that this leaves a remainder of 1 when dividing by 3.
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    (Original post by B_9710)
    Of course if you do this exercise using modular arithmetic it is very simple. If not then if a number x leaves a remainder of 2 when divided by 3 then it is of the form x=3y+2 where y is an integer.
    So x^2=(3y+2)^2=9y^2+12y+4 and you should be able to show that this leaves a remainder of 1 when dividing by 3.
    Thank you for the response but could you explain to me how you get x = 3y + 2 as I thought it would be
    x/3 +2 as the first part
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    (Original post by Zain786H)
    Thank you for the response but could you explain to me how you get x = 3y + 2 as I thought it would be
    x/3 +2 as the first part
    Because you divided it to get the answer, therefore you have to multiply to find the original number.
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    (Original post by CCauston113)
    Because you divided it to get the answer, therefore you have to multiply to find the original number.
    I understand now, thank you
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