2

A number when divided by $902$, $802$ and $702$ leaves remainder of $602$, $502$ and $402$ respectively. What would be the minimum remainder of when the number is divided by $2005$?

My attempt:

For the given number $n$, $n + 300$ is divisible by $902$, $802$ and $702$. Hence $n + 300$ can be written as $\mathrm{lcm}(902,802,702)\cdot K$ where $K$ is the integer and we can keep varying $K$ from $1$ onwards to manually check the minimum remainder. However this is not a great approach. Any simplistic approach to this problem would be appreciated.

Bill Dubuque
  • 272,048
Mathronza
  • 164
  • 2
    I think is the simplest approach. The LCM of $902,802, 702$ is not very difficult to find. Also note that both $802$ and $2005$ are divisible by $401$. – player3236 Mar 02 '21 at 15:01
  • @LearningMathematics $n + 300$ is written up like that not $n$. – Dávid Laczkó Mar 02 '21 at 19:14
  • You don't have to solve or do anything. You just need to know $n\equiv 502\pmod 802\implies n \equiv 502\equiv 101 \pmod{401} \implies n \equiv 101 + 401k \pmod {2005}$. The least remainder is $101$ if we have $n\equiv 101 \pmod 5$. And by CRT we can have that. – fleablood Mar 02 '21 at 19:58
  • A "great approach" to studying solutions of systems of congruences is to start with the CRT solvability criterion - which immediately yields the result here - see my answer. – Bill Dubuque Mar 03 '21 at 13:06
  • Did you understand the answer I posted? (the standard way to solve such problems). If not let me know what is not clear so I can elaborate. – Bill Dubuque Mar 11 '21 at 09:26

2 Answers2

1

We seek the least $\,r\in \Bbb N\,$ such that the following congruence system is solvable

$$\begin{align} &x\equiv -300\pmod{702}\\ &x\equiv -300\pmod{802}\\ &x\equiv -300\pmod{902}\\ &x\equiv\ \ \ r\quad \pmod{2005} \end{align}\qquad$$

By the CRT solvability criterion it has a solution $\!\iff\!$ each pair has a solution, and a pair $\,x\equiv a\pmod{\!m},\ x\equiv b\pmod{\!n}\,$ is solvable $\!\iff\!$ they are consistent mod the gcd $\,d=(m,n),\,$ i.e. $\,a\equiv x\equiv b\pmod{\! d},\,$ i.e. $\,\color{#c00}{d\mid a-b}.\,$ The first three are consistent since they have obvious solution $\,x = -300\,$ so it remains to check that the last is consistent with the first three, which is true for the first and third because their modulus is coprime to $2005.\,$ For the second we require $\,(2005,802)=\color{#c00}{401\mid r+300}.\,$ Thus $\,r=101\,$ is clearly the least $\,r\in\Bbb N\,$ making it solvable.

Bill Dubuque
  • 272,048
0

Lots of obfuscation.

$n \equiv 502 \pmod {802}$ so $n\equiv 502 \equiv 101 \pmod {401}$.

So $n =101 + 401K$ for some $K$ so $\pmod {2005 = 401 \times 5}$ we have $n \equiv 101 + 401k$ for $k = 0,1,2,3,4$

So $n \equiv 101 + 401k \pmod {2005}$. The least that can be is $101$ if $k=0$ and we'd have $n \equiv 101 \pmod 5$.

Now if we want

$n\equiv 602 \pmod{902}$

$n \equiv 502 \pmod {802}$

$n \equiv 402 \pmod {702}$ and

$n \equiv 101 \pmod 5$ we just need

$n \equiv 0 \pmod 2; n \equiv 602 \pmod 451; n \equiv 502 \pmod{401}; n\equiv 402\pmod{351}; $ and $n\equiv 101 \pmod 5$.

And as $2, 451, 401, 351, 5$ are relatively prime. So by CRT we can have that.

(And thank god we don't have to find that...... well, it wouldn't be that bad[1] but... we don't have to find it; it's enough to know it exists)

And so $n \equiv 101\pmod 5$ and $n \equiv 101 \pmod {401}$ and so $n\equiv 101 \pmod {2005}$ and $101$ is the smallest possible remainder.

====

[1] If we actually want to solve:

$n \equiv -300 \pmod {902, 802, 702}$ so $n\equiv -300 \pmod{2;451;401;351}$ so $n \equiv -300 \pmod {126957402}$ and $n \equiv 1\pmod 5$ so $n\equiv -300 + 126957402k \pmod {634,787,010}$ where $k = 0,1,2,3, 4$ but as $n \equiv 1 \pmod 5$ we must have $k=3$ so $n \equiv 380871906 \pmod {634,787,010}$ and indeed $380871906 \pmod {2005} \equiv 101$.

And $380871906 \equiv 602;502;402 \pmod{902;802;702}$.

fleablood
  • 124,253
  • It's not a good idea to write nonspecific critiques like "lots of obfuscation" since the OP will not have any way of knowing what you specifically refer to and as a result may be misled about on the value of remarks made by other users. – Bill Dubuque Mar 03 '21 at 02:07
  • I meant the presentation of the problem had a lot of obfuscation. It doesn't matter what any of the remainders are of any of the moduli and you really only needed to know $n\equiv 101 \pmod {401}$ – fleablood Mar 03 '21 at 05:39
  • It's still not clear what you think is obfuscated. The only obfuscation I see is that your answer greatly obfuscates the fact that the solution is an immediate consequence of the CRT solvability criterion - see my answer. Likely the problem was designed to provide exercise using this criterion. – Bill Dubuque Mar 03 '21 at 13:09
  • Right. But the question is about $\pmod 2005 = 401\times 5$ so we only need concern ourselves the the congruences with modulus not relative prime to $401$ and $5$ and that is only $802$. Then $\mod 702,902$ congruences are nearly irrelevent (the only thing that matters about those is that $2$ is the only reused modulus and that these are consisten mod $2$ and that's only relevent as the question [not anything the student would do] would be invalid if they weren't. It looks like the quesion will have a lot to do with a lot of information but it uses very little. – fleablood Mar 03 '21 at 15:49