# Remainders with specific divisors ## 18 thoughts on “Remainders with specific divisors”

1. dushyant133 |

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. catcracker |

Hi Dushyant,

That was actually intended to be just an exclamation point (and not a factorial symbol). Sorry for the confusion…

J

• dushyant133 |

Thanks 🙂

3. Aniruddh |

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.

• sharan |

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

• catcracker |

I think you have the number wrong – it is an 8-digit number and we are measuring two digits at a time from the back.

regards
J

• Shraddha |

for the second example how did you get the ans as 90?

• catcracker |

Not sure which one you mean? I haven’t got the answer as 90 for any of the examples I gave…

regards
J

• catcracker |

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

4. catcracker |

Anirudh, your approach is fine. Digits will be counted from the decimal point always.

regards
J

5. kooks |

Hi sir, In one of the questions above.. 19^21 is of the form 4k+3.. isn’t it 4k+(3^21) ?

• catcracker |

Think of it as (20-1)^21 = (4k-1)21 = 4m + (-1)^21 = 4m – 1 = 4n + 3.

regards
J

• Alia |

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?

• catcracker |

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

6. kooks |

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 🙂

• catcracker |

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

7. Manish |

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.

• catcracker |

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