6

A question says:

"Which $n \in \Bbb N$ should the numbers $1991$, $1805$, $2146$ be divided into, so that there will be the same remainders left?"

I arranged them in order: $1805,1991,2146.$ Then did subtraction: $1991-1805=186,$ $2146-1991=155,$ then $186-155=31.$ This last result is the answer. But I am skeptical about my way of finding the answer (and which seems to be a little bit stranger to me), as it wouldn't work in some cases. I would like 1) to know your thoughts on my method, 2) you to provide some hints in order to show the right way of solving this.

user36339
  • 575
  • 7
  • 23

2 Answers2

4

If the remainders from $1991$ and $1805$ are the same, then these numbers are congruent mod $n$, so $n$ divides $1991-1805=186$. Similarly, $n$ must divide $2146-1991=155$. So you need a number that divides both $186$ and $155$, so it will divide their difference: $186-155=31$. Since $31$ is prime its only divisors are $31$ and $1$, and technically the answer is both of these.

So your method is valid. And very fast, so I wouldn't expect anything faster.

2'5 9'2
  • 54,717
4

Given three numbers $a,b,c$ your are looking for $\gcd (b-a,c-b,c-a)$. Your approach is a good step and the answer is guaranteed to divide $b-a$ and $c-b$. You have done one step in Euler's method, but the answer could be smaller. For example, let the numbers be $1,19,33$. You would do $19-1=18, 33-19=14, 18-14=4$ but are not done as the $\gcd$ is $2$. You can now keep going with $14-3 \cdot 4=2$ and you are done because $2$ divides both $18$ and $14$.

Ross Millikan
  • 374,822
  • Sorry for misunderstanding, do you want to say that 2 is a number that when divides 1, 19 and 33 produces the same remainders, here we have 19=29+1, 33=216+1 but 1 can't be divided into 2 and produce the common remainder 1. Could you shed light on this question? – user36339 Sep 09 '17 at 05:35
  • 1
    Yes, $1=0\cdot 2 + 1$. It leaves remainder $1$ when divided by $2$. – Ross Millikan Sep 09 '17 at 05:39
  • Wonderful! I would give V both answers (and one below). – user36339 Sep 09 '17 at 05:42