The inverse of $a$ modulo 39 is $b$. What is the inverse of $4a$ modulo 39 in terms of $b$? Give your answer as an expression in terms of $b

$$\small{\text{ The inverse of $a$ modulo 39 is $b$. }}\\ \small{\text{ What is the inverse of $4a$ modulo 39 in terms of $b$? }}\\ \small{\text{ Give your answer as an expression in terms of $b$. }}$$

$$\small{\text{ $a\cdot b \equiv 1 \pmod{39}$ and $4a\cdot x \equiv 1 \pmod{39}$ so $ab=4ax$ and $x=\frac{1}{4}b$ }}\\ \small{\text{ The inverse of $4a \pmod{39}$ is $\frac{1}{4}b$ }}$$

That is not correct, because if b is not a multiple of 4, then the inverse would be a fraction, and division is not allowed in modular arithmetic.

So let's say the new inverse is xb. 

Then we would have 4a*xb ==1 (mod 39). We will have 4x*ab (mod 39)

We previously know that ab (mod 39)=1, so we can reduce that.

We have the 4x ==1 (mod 39). We need to find x. 

The inverse of 4 (mod 39) is 10, so we have x is 10.

Therefore, the answer is 10b.

Hope this helps :)