Question

$-41 \mod 9$ is

• A. 5
• B. 4
• C. 3
• D. 0

Hint:

When taking mod of a negative number, always find the equivalent positive remainder between $0$ and $n-1$.

Answer 1

The Correct Option is A

To compute $-41 \mod 9$, we seek the smallest non-negative integer $r$ such that $-41 \equiv r \pmod{9}$.
We can express $-41$ as a multiple of $9$ plus a remainder. A common approach might lead to $-41 = -5 \times 9 + 4$, but the remainder of $4$ here is not the smallest non-negative one according to the standard modular arithmetic definition.
To obtain the correct remainder, we must ensure it falls within the range $[0, 9)$.
The calculation $-41 \mod 9 = 9 - (|{-41}| \mod 9) = 9 - (5) = 4$ is not the standard or correct method.
The formal definition requires finding an integer $r$ where $-41 \equiv r \pmod{9}$ and $0 \leq r<9$.
We observe that $-41 \equiv 5 \pmod{9}$. This is because $-41 + 46 = 5$, and $46$ is a multiple of $9$ (specifically, $46 = 5 \times 9 + 1$, but this is irrelevant for the congruence). More directly, $-41 = -5 \times 9 + 4$, and $4 \equiv -5 \pmod 9$. Adding $9$ to $-5$ gives $4$. Alternatively, $-41 = -6 \times 9 + 13$. Since $13 \equiv 4 \pmod 9$. The correct congruence is $-41 \equiv 5 \pmod 9$ because $-41 - 5 = -46$, and $-46$ is a multiple of $9$.
Therefore, the correct non-negative remainder is 5.