How can I solve the following equation in modular arithmetic?

Question

I had a test on modular arithmetic. One of the questions was

Find x such that $$ x * 11 = 5 \mod 64$$ or determine that there is no such solution.

How can I solve such an equation? I know from calculations in calc that the solution is

$$ x = 175$$

However how can I solve this in a smart way?

Find the multiplicative inverse of $11$ mod $64$ and multiply by $5$ — Brendan Sullivan, Jul 09 '21 at 06:40
"the solution is $x=175$" $-$ That's just one solution out the family of solutions $,47 + 64,k,$. — dxiv, Jul 09 '21 at 06:43
Or we can use Newton's method (Hensel's Lemma) to lift an easy inverse mod $8$ to mod $8^2$ $$\color{#0a0}{3}^{-1}\equiv \color{#c00}3\pmod{!8},\Rightarrow, \bmod 8^2!:\ \dfrac{5}{\color{#0a0}3!+!8} \overset{\large \times\ \color{#c00}3_{\phantom{|}}}= \dfrac{15}{1+4(8)} = 15(1!-!4(8)) = 15!-!(-4)8 = 47$$ — Bill Dubuque, Jul 09 '21 at 07:43

score 1 · Accepted Answer · answered Jul 09 '21 at 07:10

1

Here's the general approach. Using the Extended Euclidean Algorithm with $64$ and $11$, you find that $$11\cdot (-29)+64\cdot 5=1.$$ Thus $$11\cdot (-29)\equiv1\mod 64.$$ Coming back to your equation, we may multiply both sides by $-29$ to get $$x\cdot 1\equiv -145\mod 64.$$ As $$-145\equiv 47\mod 64,$$ we conclude that $x\in\{47+64k\mid k\in\mathbb Z\}$.

answered Jul 09 '21 at 07:10

Zuy

4,656

Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 09 '21 at 07:20
Please note that the question didn't have much context, so you should have only posted an hint, and not a full solution. – ultralegend5385 Jul 09 '21 at 07:25
Thank you for these remarks. – Zuy Jul 09 '21 at 08:05

Stephen Donovan · Answer 2 · 2021-07-09T06:58:15.253

There's probably a better way of doing this, but this was my approach.

$11x \equiv 5 \pmod{64}$ implies that there exists some integer $a$ such that $11x = 64a + 5.$ Because of this we must have $64a \equiv -2a \equiv -5 \pmod{11} \Leftrightarrow a \equiv 8 \pmod{11}.$ Letting $a = 8$ gives us $11x = 64(8) + 5 = 517 = 47\cdot11.$ So, any $x \equiv 47 \pmod{64}$ will work. (and $x = 175$ is in this class because $175 = 2(64) + 47$)

Edit: I realized this is actually a bit circular since I relied on the solution to $2a \equiv 5 \pmod{11}$ but I think this division is pretty trivial since $16 = 5 + 11 = 2\cdot8.$ In any case you could repeat the process to solve this equation: $2a = 5 + 11b \Leftrightarrow b \equiv 1 \pmod{2}, 2a = 5+11 \Leftrightarrow a = 8.$

Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. — Bill Dubuque, Jul 09 '21 at 07:21

How can I solve the following equation in modular arithmetic?

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