1
$\begingroup$

This is the work I have tried somehow I keep getting 4 instead of 11:

$26 = 1\times 19 + 6$

$19 = 3\times 6 + 1$

Now,

$1= 19 - 3\times 6$

Sub $6 = 26 - 19$

$1 = 19 -3(26-19)$ $1 = -3\times 26 + 4\times 19$

Therefore multiplicative inverse is somehow 4

$\endgroup$
  • 5
    $\begingroup$ 26=1*19+7 actually. $\endgroup$ – Roddy MacPhee Oct 5 at 11:43
  • $\begingroup$ Oh my gosh thank you so much! I've been looking at this problem for like 2 hours trying to figure out where my error was I was even able to figure out the more complex problems. Thank you so much $\endgroup$ – user3371137 Oct 5 at 11:46
  • 3
    $\begingroup$ What do you mean by the multiplicative inverse? The multiplicative inverse is the number you multiply by another number to get $1$. Therefore, the multiplicative inverses of $19$ and $26$ are $\frac{1}{19}$ and $\frac{1}{26}$, respectively. $\endgroup$ – John Douma Oct 5 at 11:48
  • 3
    $\begingroup$ @John, I suspect what user has in mind is the multiplicative inverse of $19$, working modulo $26$. $\endgroup$ – Gerry Myerson Oct 5 at 11:55
  • 1
    $\begingroup$ You've not asked a proper question here. I see a modular arithmetic tag. What is your modulus here, for example? $\endgroup$ – Allawonder Oct 5 at 12:08
0
$\begingroup$

You are searching for an integer $a$, such that $19×a\equiv 1$ (mod $26$), i.e. $19a=26k+1$ (where $k$ is an integer also).

So, $a= \frac {26k+1}{19}$ so just try some values of $k$ until you have a first integer. You'll have it first at $k=8$, $(26(8)+1)/19= 209/19=11$ and that is your inverse.

Note that when solving with (mod $26$) then you know that $a<26$ so you only try some $k$'s here that are $1\le k\le 18$.

If for all these $k$'s we didn't an integer then $19$ wouldn't have had an inverse (but thankfully here it did).

$\endgroup$
  • $\begingroup$ you can skip all odd k as 7*k+1 will be even. $\endgroup$ – Roddy MacPhee Oct 5 at 12:57
  • $\begingroup$ Yes you're right, I just wanted to point out something that he can always use. But I don't understand what is the downvote for? $\endgroup$ – Fareed Abi Farraj Oct 5 at 13:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.