# Remainders – 2

## 13 thoughts on “Remainders – 2”

1. Gaurav |

Dear Mr J,

Why is the common remainder to be found in the condition of two numbers, in the range – 0 to 1 less than the LCM?

• 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

2. Gaurav |

* 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 ?

3. Sandipan |

Sir,
That “Theorem” stated in the last line should also mention x never equals y…should it not?

• It is not a theorem really, Sandipan – just a minor result by observation ๐ 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

4. Sandipan |

ok sir…

5. Gaurav Chaudary |

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

6. Gaurav Chaudary |

Thank You ๐

7. seema |

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..

• See the next two posts (Remainders -3 and -4) With those and the idea of an LCM/HCF you should be able to handle question of this type.

regards
J

8. Anshuman Shaw |

Sir thereโs a small typo it should be 7*3 + 3