11 thoughts on “Remainders – 2

    • Let’s say one number (condition) repeats every 7 steps and another every 5 steps. Then they will both repeat after every 35 steps. Now if I am saying that in every 35 consecutive numbers there will be exactly 1 solution, isn’t that equivalent to saying that there will be one solution between 0 and 34 (which include 35 consecutive numbers)?

      regards
      J

  1. * my bad – not common remainder,

    but for two numbers say 2k+7 and 3k+5, why is the number which satisfies both the criteria in the range from 0 to LCM-1 ?

    • It is not a theorem really, Sandipan – just a minor result by observation 🙂 And if x equals y, then it will still hold good trivially as the first such number will then be x itself (which is still between 0 and [LCM(A, B) – 1]). For example if the remainder with 5 = 2 and with 7 = 2, then the first such number will be 2 and subsequent numbers will be 2 + 35, 2 + 70 and so on.

      regards
      J

  2. In the last question –

    “Find the number which leaves remainders 2 and 3……” ,

    why does the approach of taking LCM of 6m, 11n and 5k together go wrong?

    • It doesn’t go wrong as such, just that you would have to check all numbers up to 330. This saves time – taking 2 at a time, and then taking the combined result with the 3rd.

      regards
      J

  3. A number when divided by 33 leaves remainder 4, what will be the no. of possible remainders when it is divided by 55? please give a generic idea of such questions..

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