2÷5=𝚝(𝚏𝚒𝚗𝚒𝚝𝚎 7).𝙵𝚒𝚗𝚍 𝚝𝚑𝚎 𝚟𝚊𝚕𝚞𝚎 𝚘𝚏 𝚝

The notation "finite 7" indicates that we are working in modular arithmetic modulo 7, often denoted as $\mathbb{Z}_7$. The equation $2 \div 5 = t$ means we need to find $t$ such that $2 \equiv 5t \pmod{7}$.

In modular arithmetic, division by a number $a$ is equivalent to multiplication by its multiplicative inverse $a^{-1}$, where $a \cdot a^{-1} \equiv 1 \pmod{7}$. So, $t \equiv 2 \cdot 5^{-1} \pmod{7}$.

We need to find an integer $x$ such that $5x \equiv 1 \pmod{7}$. We can test values for $x$ from 0 to 6:

• $5 \times 0 = 0 \pmod{7}$
• $5 \times 1 = 5 \pmod{7}$
• $5 \times 2 = 10 \equiv 3 \pmod{7}$
• $5 \times 3 = 15 \equiv 1 \pmod{7}$ Thus, the multiplicative inverse of 5 modulo 7 is 3, i.e., $5^{-1} \equiv 3 \pmod{7}$.

Substitute the inverse back into the equation for $t$:
$t \equiv 2 \cdot 3 \pmod{7}$
$t \equiv 6 \pmod{7}$

The value of $t$ is: