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)?
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.
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.
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..
CAT-holics 
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
* 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 ?
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 🙂 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
ok sir…
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
Thank You 🙂
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
Sir there’s a small typo it should be 7*3 + 3
Right you are! My bad…
regards
J