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 $b$.
$$\small{\text{
The inverse of $a$ modulo 39 is $b$.
}}\\
\small{\text{
What is the inverse of $4a$ modulo 39 in terms of $b$? }}\\
\small{\text{
Give your answer as an expression in terms of $b$.
}}$$
$$\small{\text{
$a\cdot b \equiv 1 \pmod{39}$ and $4a\cdot x \equiv 1 \pmod{39}$ so $ab=4ax$ and $x=\frac{1}{4}b$
}}\\
\small{\text{
The inverse of $4a \pmod{39}$ is $\frac{1}{4}b$
}}$$
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$$\small{\text{
The inverse of $a$ modulo 39 is $b$.
}}\\
\small{\text{
What is the inverse of $4a$ modulo 39 in terms of $b$? }}\\
\small{\text{
Give your answer as an expression in terms of $b$.
}}$$
$$\small{\text{
$a\cdot b \equiv 1 \pmod{39}$ and $4a\cdot x \equiv 1 \pmod{39}$ so $ab=4ax$ and $x=\frac{1}{4}b$
}}\\
\small{\text{
The inverse of $4a \pmod{39}$ is $\frac{1}{4}b$
}}$$
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.