18 thoughts on “Remainders with specific divisors

  1. In solution to “Find the remainder when 14285 is divided by 9 ?”, it is mentioned in last line “this is the basis for the divisibility test for 9!” what does this mean ? I believe 9! is not for trailing 9s.

  2. Dear Sir,

    For finding the remainder with 99, what happens when there are odd no. of digits? We must consider the first digit separately and then take two digits at a time, right?

    For example Rem.(1728133/99) = (1 + 72 + 81 +33)/99 = 88

    Also, Rem. (123123123….99 digits)/99 = 1 + [(23 + 12 + 31) + (23 + 12 + 31) +….16 times] +23 = [24 + 66X16] /99 = 90

    Is my approach correct? Is there an easier way? Please confirm.

    • hi,
      for the first case when u add 1,72,81 and 33 you get 187. isn’t the answer 87.
      i a little confused please correct me if i am wrong.

      thanks

      • If you mean the example posted by Aniruddh?

        As he has shown, it comes to [24 + 66X16] /99. At this point we can either compute the whole thing, or realise that 66 and 99 have a good synergy; thus we can write it as [24 + 66 + 66*15] / 99. Now 66 * 15 is divisible by 99, so the remainder will just be given by 24 + 66 which is 90.

        regards
        J

      • Hi… Can you elaborate on how you reached 4n+3. Also why have we expanded 20-1 as 4k-1 and not 5k-1 since we are dividing by 5?

      • For your first query – a multiple of 4 minus 1 will be the previous multiple of 4 plus 3. Think about it….take a few small numbers and check.

        For your second – the cycle we got was of 4 steps. So the pattern repeats every 4 steps and we need to find how many extra steps are there over a multiple of 4 i.e. the remainder when dividing by 4…

        regards
        J

  3. Hello Sir.. Thanks for reply.. Sir while solving questions on similar lines, i have observed that in most of the cases the binomial expression reduces to same structure as how it is before opening using binomial theorem.. like 97^97 = (10n + 7)^97 is of the form 10n + 7. And 56^52 = (10k + 6)^52 is of the form 10k + 6. Is this a general form/result that i can use without giving second thoughts ? And i ll be very grateful if you can explain for any one of the aforementioned cases also ?

    Thanks a lot 🙂

    • No you cannot. The operative word is most (but not all). You can’t figure out which ones WON’T give that pattern. The two examples you chose were coincidences – the first one, 7 has a cyclicity of 4 and hence when raised to 97 (4k+1 form) gives the original value again. The second, well, any power of a number ending in 6 will also end in 6. Often the examiner might choose such a number because it is easier to set, but you cannot assume it!

      (See https://cat100percentile.com/tag/last-digit/ if you need to brush up the idea of cyclicity.)

      regards
      J

  4. how to we break the sequence??
    like how can 12345123451234……200 times be equal to (12+34+51+23+45)*20 times..
    please tell the the portion I need to look into to get clear idea of this part.

    • No portion 😛 Just logic. For remainder with 99 we have to break it 2 digits at a time. If we keep doing this, after a while we get the sequence repeating (12, 34, 51, 23, 45, 12). At that point we stop and check how many times it repeats. Simple as that.

      regards
      J

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