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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • S Offline
      shin88
      last edited by

      Pls help me on the following Qin. Thanks

      If 600 = 2^3 x 3 x 5^2. Find the smallest value of p such that 600p is a perfect cube.

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      • J Offline
        jieheng
        last edited by

        shin88:
        Pls help me on the following Qin. Thanks

        If 600 = 2^3 x 3 x 5^2. Find the smallest value of p such that 600p is a perfect cube.
        600 = 2^3 x 3 x 5^2

        To be a perfect cube , the index of the prime factors of 600p must be multiple of 3 ,

        600p = 2^3*3^3*5^3

        p = 3^2*5 = 45

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        • S Offline
          shin88
          last edited by

          Thanks for your solution. I understand how u get 2^3x3^3x5^3 but how to get 3^2x5 ?

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          • J Offline
            jieheng
            last edited by

            shin88:
            Thanks for your solution. I understand how u get 2^3x3^3x5^3 but how to get 3^2x5 ?

            600 = 2^3 x 3 x 5^2

            600p = 2^3 * 3^3 * 5^3
            2^3 * 3 * 5^2 * p =2^3 * 3^3 * 5^3

            p = 3^2*5 = 45

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            • S Offline
              shin88
              last edited by

              Thanks a lot, I finally understand :rahrah:

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              • 2 Offline
                2DMommy
                last edited by

                Please help with this q :


                Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (eg 123123). The number formed will always be divisible by 7,11 and 13.
                A) explain why
                B) what are the other factors of any such number formed ?

                Thks !

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                • W Offline
                  WTK
                  last edited by

                  Help please!


                  Find the greatest number that will divide 171, 255 and 304 so as to leave the same remainder in each case.

                  Kindly show the workings. Thank you.

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                  • J Offline
                    Jtutor
                    last edited by

                    2DMommy:
                    Please help with this q :


                    Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (eg 123123). The number formed will always be divisible by 7,11 and 13.
                    A) explain why
                    B) what are the other factors of any such number formed ?

                    Thks !
                    Hi 2DMommy,

                    A) 7x11x13 = 1001
                    1001x100 = 100100 (answer)
                    B) Other factors are 2^2 x 5^2

                    Hope the above is correct.

                    Cheers!
                    Jtutor

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                    • MathIzzzFunM Offline
                      MathIzzzFun
                      last edited by

                      WTK:
                      Help please!


                      Find the greatest number that will divide 171, 255 and 304 so as to leave the same remainder in each case.

                      Kindly show the workings. Thank you.
                      Let the greatest number be n, and the remainder be R

                      171 = An + R
                      255 = Bn + R
                      304 = Cn + R
                      where A,B,C are integers

                      (B-A)n = 255 - 171 = 84 = 12 x 7
                      (C-B)n = 304 - 255 = 49 = 7 x 7
                      (C-A)n = 304 - 171 = 133 = 19 x 7

                      n = 7

                      The greatest number is 7

                      cheers.

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                      • W Offline
                        WTK
                        last edited by

                        Thanks MathIzzzFun.

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