I feel like I am missing something really obvious here but I can't explain why these two are equivalent. I understand that $5t \equiv 1 \pmod 6$ but I can't make the next step.
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well first we subtract $1$ from either side, to get $5t\equiv1\pmod6$. To isolate $t$ we want to 'undo' the multiplication by $5$, so we multiply by its multiplicative inverse $5^{-1}$. In this case, recognize $5\equiv-1\pmod6$ and $5^2\equiv1\pmod6$ so $5$ is its own multiplicative inverse... – obataku Aug 17 '13 at 14:10
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Starting with $$5t \equiv 1 \pmod 6,$$ if we add $5+t$ both sides, we obtain $$6t+5 \equiv 6+t \pmod 6,$$ or equivalently $$t \equiv 5 \pmod 6.$$
Douglas S. Stones
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If you were solving the equation $5t = 1$ you'd multiply both sides by $5^{-1} = \frac{1}{5}$ which is the number which satisfies $5^{-1}.5 = 5.5^{-1} = 1$. Then $t$ would be isolated on the left hand side.
When solving such an equation modulo $m$, the approach is the same. The interpretation of $5^{-1}$ is the same (it undoes multiplication by $5$) but what $5^{-1}$ is equal to is different. In this case, $5^{-1}$ is the unique element of $\{0, 1, \dots, m-1\}$ such that $5.5^{-1} = 5^{-1}.5 \equiv 1 \pmod m$ - such an element exists if and only if $5$ and $m$ have no common factors (i.e. they are coprime).
To solve the equation $5t \equiv 1 \pmod 6$, you need to find what $5^{-1}$ is, keeping in mind that $5^{-1} \in \{0, 1, \dots, 5\}$.
Michael Albanese
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1Ok do we multiply each side by 5 so that we have 25t $\equiv$ 5(mod6) which is equivalent to 1$.$t $\equiv$ 5(mod6) ? – Kom Aug 17 '13 at 14:06
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1Yes. Another way you can think about it is that $t$ must be $5^{-1}$ then use the fact that $5\times 5 \equiv 1 \pmod 6$. However, the way you described it is the best way to think about it as it works in general. – Michael Albanese Aug 17 '13 at 14:14