Use Euclid's algorithm to find a multiplicative inverse of 15 mod 88, hence solve the linear congruence $15x\equiv 20 \pmod{88}$

So far I have:



Backwards substitution gives





now very stuck what to do from here:(

  • $\begingroup$ Where did the $18$ come frome? $\endgroup$ – Thomas Andrews Dec 4 '16 at 15:07
  • 2
    $\begingroup$ And $1\neq 15-1\cdot 13$... $\endgroup$ – Thomas Andrews Dec 4 '16 at 15:07

This is much simpler if you use the extended Euclidean algorithm in fraction form

$${\rm mod}\ 88\!:\ \ x\equiv \dfrac{20}{15}\equiv \dfrac{4}3\equiv \dfrac{-84}3\equiv -28\equiv 60$$

Remark $\ $ It's valid to cancel $\,5\,$ above since $\,5\,$ is coprime to $\,88\,$ so invertible, so cancellable, i.e.

$$\begin{align} \color{#c00}5\cdot 3x\equiv \color{#c00}5\cdot 4\!\! \overset{\large \ \ \,\times\ \color{#c00}{5^{\large -1}}} \iff&\ 3x\equiv 4\\[.4em] {\rm so}\quad x\equiv \dfrac{20}{15}\ \iff&\ \ x\equiv \dfrac{4}3\end{align}$$

Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.


You have missed a few steps. Also I'm not sure where the number $18$ came from in $18w$.

Normally in the initial part you should continue until you get to 1. E.g.

$88=5\times15+13$ hence $13=88-5\times13\tag{1}$

$15=1\times13+2$ hence $2=15-1\times13\tag{2}$

$13=6\times2+1$ hence $1=13-6\times2\tag{3}$

Then work backwards from there:

$1=13-6\times2$ using $(3)$

$1=13-6\times(15-1\times13)=7\times13-6\times15$ using $(2)$

$1=7\times(88-5\times15)-6\times15=7\times88-41\times15$ using $(1)$

Hence $-41\times15\equiv1\pmod{88}$

Or written with a positive number: $47\times15\equiv1\pmod{88}$

So $20\times47\times15=940\times15\equiv20\pmod{88}$

$940=10\times88+60$ hence $60\times15\equiv20\pmod{88}$


To solve $15x\equiv 20\pmod{88}$, subtract consecutive equations to find $15^{-1}\bmod 88$ first:

$$88=(1)(88)+(0)(15)\\ 15=(0)(88)+(1)(15)\\ 13=(1)(88)+(-5)(15)$$

(because $88-5\cdot 15=13$)



Therefore $(15)(-41)\equiv 1\pmod{88}$, so $$15^{-1}\equiv -41\equiv 47\pmod{88}$$

Therefore $$x\equiv 20\cdot 15^{-1}\equiv 20\cdot 47\pmod{88}$$

$$\equiv 940\equiv 60\pmod{88}$$


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