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(a) $x+40 ≡ 1 \pmod {99}$.

(b) $40x ≡ 1 \pmod {99}$.

Is the answer for a $x \equiv -39 \pmod {99}$ and b $x\equiv\frac{1}{40}\pmod {99}$?

I believe I am doing it correctly.

Juniven Acapulco
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modlim
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    Right, but $\dfrac 1 {40}$ makes no sense unless you express it as an integer $n$ such that $40n\equiv 1 \mod 99$. – Pedro Sep 23 '13 at 23:27
  • How to simplify it? – modlim Sep 23 '13 at 23:36
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    Euclidean algorithm, or $$\frac{1}{{40}} = \frac{{99 + 1}}{{40}} = \frac{{100}}{{40}} = \frac{5}{2} = \frac{{99 + 5}}{2} = \frac{{104}}{2} = 52$$ – Pedro Sep 23 '13 at 23:40

1 Answers1

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$(a)\quad x+40 ≡ 1 \pmod{99} \iff x \equiv -39 \pmod {99}\iff x \equiv -39 + 99 = 60 \pmod {99}$.

$(b) x \cdot 40 ≡ 1 \pmod {99} \iff x \equiv \frac 1{40} \pmod {99}\quad?$

Well, sort of. We do need to find the multiplicative inverse of $40$, but $\mod 99$.

As Pedro shows in the comments, we can find the multiplicative inverse of $40 \mod 99$ by using the Euclidean Algorithm, or by using the following algorithm:

$$\frac{1}{{40}} = \frac{{99 + 1}}{{40}} = \frac{{100}}{{40}} = \frac{5}{2} = \frac{{99 + 5}}{2} = \frac{{104}}{2} = 52$$

Note, we can check to ensure we've found the multiplicative inverse of $40 \mod 99$: $$40 \cdot 52 = 2080\equiv 1 \pmod{99}$$ $$\begin{align}\implies (b)\;\; x \cdot 40 \cdot 52 &\equiv 1 \cdot 52 \pmod {99} \\ \\ x & \equiv 52 \pmod{99}\end{align}$$

amWhy
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