![]() The same sort of argument works for all the numbers $2$ to $10$. So the remainder when $-1$ is divided by $6$ is $5$. The remainder when the number $a$ is divided by $6$ is the integer $r$ such that $0\le r\le 5$, and there is some integer $q$ such that $a=6q+r$.Īpply this definition to $a=-1$. The congruence notation is quite helpful here, but we will try to do without. So $x=-1$ works for every one of our moduli.Īdded details: There is something not very intuitive about the remainder when a negative integer like $-1$ is divided by, say, $6$. We can restate these congruences in the form $x$ is congruent to $-1$ modulo $2$, $x$ is congruent to $-1$ modulo $3$, $x$ is congruent to $-1$ modulo $4$, and so on up to $x$ is congruent to $-1$ modulo $10$. We are told that a certain integer $x$ is congruent to $1$ modulo $2$, congruent to $2$ modulo $3$, congruent to $3$ modulo $4$, and so on up to congruent to $9$ modulo $10$. What's going on is easier to see if we use congruence notation. Its only flaw is that it is not positive, but that flaw can be dealt with. In our particular case, finding an integer that works was easy, since $-1$ is clearly a solution. A specified sequence of remainders may not be achievable, and even when it is, efficiently finding an integer that works involves using the Chinese Remainder Theorem, or some equivalent procedure. Remark: This type of problem is in general much more messy to solve. So the smallest positive integer that works is $M-1$. The integers that have the desired property are all the integers of the form $-1+kM$, where $M$ is the least common multiple of the integers $2,3,4,\dots, 10$ and $k$ ranges over the integers. But there is a smallest positive integer. ![]()
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