# Multiplicative Inverse of $19 \pmod{26}$ [closed]

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

• 26=1*19+7 actually. – Roddy MacPhee Oct 5 at 11:43
• 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 – user3371137 Oct 5 at 11:46
• 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. – John Douma Oct 5 at 11:48
• @John, I suspect what user has in mind is the multiplicative inverse of $19$, working modulo $26$. – Gerry Myerson Oct 5 at 11:55
• You've not asked a proper question here. I see a modular arithmetic tag. What is your modulus here, for example? – Allawonder Oct 5 at 12:08

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).