1
$\begingroup$

I am looking for a number that when multiplied by any number and divided by 10000 never leaves the 3 digit number as 291 , I mean I am looking for a number that leaves a remainder as 1231 , 1001 ...etc and not like 0012,0291,0001.

  • 0
    I am not sure about what you are asking. Are you looking for a number $N$ such that the division algorithm gives $Nk=10000q+r$ with $1000\leq r\leq9999$ for all $k>0$ ? As stated, this is impossible: for any $N$ take $k=10000$.2012-07-23
  • 0
    I am asking a number say "n" which when multiplied by a number say 1221 and divided by 10000 the remainder must be a four digit remainder like 1234 and not like 0001 (which is a single digit number)2012-07-23
  • 0
    As I said, this is impossible unless you give some restrictions on the "multiplier" (like $1221$ in your comment)2012-07-23
  • 2
    You can break this into two cases: 1) $gcd(n, 10,000) = 1$ and 2) $gcd(n, 10,000) \neq 1$. In the second case, you can clearly see that for some $k,\ kn = 0 (\text{mod}\ 10,000)$. In the first case, there will be some $k$ with $kn = 1 (\text{mod}\ 10,000)$. So, no such $n$.2012-07-23
  • 1
    So you won't accept remainder 0 either? Otherwise numbers like $1000$, $1250$, $2500$, $5000$, or $10000$ itself would work. But it is clearly impossible to not get zero. Multiply by $10000$ :-)2012-07-23
  • 0
    lol, my first impulse was to suggest $5,000$2012-07-23

2 Answers 2

2

We can say even more. By Bézout's identity if your number $k$ is coprime to $10000$ (so doesn't have any factors of 2 or 5) we can find $n$ to multiply it by so the product ends in $0001$, leaving a remainder of $1$ when you divide by $10000$. That is, we can find $m$ such that $nk=m10000+1$. If $k$ has factors of $2$ or $5$ we can do the same with the $2-5$ part of $k$ replacing $1$. So if $k=2920$, it has a factor $80$ and we can find $n$ such that $nk=m10000+80$. In fact $24*2920=70080$

1

Such a number does not exist for the next reason: $(a*10000)=b$ for $ a,b\in \mathbb N$. however if you discard cases where the remainder is zero. numbers of the form $a*1000$ will work.