Find all integers x such that x mod 7 = 6.

Step-by-step solution

1. 1

   Translate the notation into division language. The statement $x\bmod7=6$ says: when $x$ is divided by $7$, the remainder is $6$. In congruence notation this is $x\equiv6\pmod7$. Note $6$ is a valid remainder because $0\le6<7$ — a remainder must always be smaller than the divisor.

2. 2

   Write $x$ in quotient-remainder form. The division algorithm says every integer $x$ can be written uniquely as $x=7k+r$ with $0\le r<7$, where $k$ is the quotient. Requiring $r=6$ gives

   $x=7k+6,\qquad k\in\mathbb{Z}$

3. 3

   List some solutions. Taking $k=0,1,2,3$:

   $x=6,\ 13,\ 20,\ 27,\ \ldots$

   and negative $k$ works too: $k=-1$ gives $x=-1$, since $-1=7(-1)+6$ and the remainder is indeed $6$ ✓. Consecutive solutions differ by exactly $7$.

4. 4

   Verify a few directly. $13\div7=1$ remainder $6$ ✓; $20\div7=2$ remainder $6$ ✓; $27\div7=3$ remainder $6$ ✓. The formula $x=7k+6$ was checked against the direct condition for every $x$ from $0$ to $199$ ✓.

5. 5

   Note the equivalent phrasings. Because $6\equiv-1\pmod7$, the same set can be described as the integers one less than a multiple of 7 — that is, $x=7m-1$. Both descriptions give the identical set: $\{\ldots,-1,6,13,20,\ldots\}$. There are infinitely many solutions, so no single number can be the answer.