Compute the multiplicative inverse of $201$ modulo $299$. Express your answer as an integer from $0$ to $298$.
Let $a$ be the inverse of $201$ modulo $299$. Then, by the definition of the inverse, $201\cdot a \equiv 1\pmod{299}$. We are looking for an integer $a$ that satisfies this congruence.

To make our task easier, we note that $603\equiv 5\pmod{299}$, and so \begin{align*}
603\cdot 60 &\equiv 5\cdot 60 \\
&= 300 \\
&\equiv 1\pmod{299}.
\end{align*}Now we write $603$ as $201\cdot 3$: $$201\cdot 3\cdot 60 \equiv 1\pmod{299}.$$Thus, the inverse we seek is $a = 3\cdot 60 = \boxed{180}$.
180
Level 5
Number Theory
Problem: Compute the multiplicative inverse of $201$ modulo $299$. Express your answer as an integer from $0$ to $298$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the multiplicative inverse of $201$ modulo $299$, we need to determine an integer $x$ such that $201x \equiv 1 \mod 299$. This requires that $201$ and $299$ are coprime, meaning their greatest common divisor (gcd) must be $1$. 

We can use the Extended Euclidean Algorithm to find the gcd and the coefficients that express this gcd as a linear combination of $201$ and $299$. 

First, we apply the Euclidean algorithm:
1. Compute $299 \div 201$, which gives a quotient of $1$ and a remainder of $98$. Thus, $299 = 201 \cdot 1 + 98$.
2. Next, compute $201 \div 98$, yielding a quotient of $2$ and a remainder of $5$. Therefore, $201 = 98 \cdot 2 + 5$.
3. Then, compute $98 \div 5$, resulting in a quotient of $19$ and a remainder of $3$. So, $98 = 5 \cdot 19 + 3$.
4. Next, compute $5 \div 3$, which gives a quotient of $1$ and a remainder of $2$. Thus, $5 = 3 \cdot 1 + 2$.
5. Then, compute $3 \div 2$, yielding a quotient of $1$ and a remainder of $1$. Therefore, $3 = 2 \cdot 1 + 1$.
6. Finally, compute $2 \div 1$, which gives a quotient of $2$ and a remainder of $0$. Thus, $2 = 1 \cdot 2 + 0$.

Since the last non-zero remainder is $1$, we confirm that $\gcd(201, 299) = 1$, indicating that an inverse exists.

Next, we backtrack using the Extended Euclidean Algorithm to express $1$ as a linear combination of $201$ and $299$:
1. From $1 = 3 - 1 \cdot 2$.
2. Substitute $2$ from $5 = 3 \cdot 1 + 2 \Rightarrow 2 = 5 - 3$ into the equation: $1 = 3 - 1 \cdot (5 - 3) = 2 \cdot 3 - 1 \cdot 5$.
3. Substitute $3$ from $98 = 5 \cdot 19 + 3 \Rightarrow 3 = 98 - 19 \cdot 5$: $1 = 2 \cdot (98 - 19 \cdot 5) - 1 \cdot 5 = 2 \cdot 98 - 39 \cdot 5$.
4. Substitute $5$ from $201 = 98 \cdot 2 + 5 \Rightarrow 5 = 201 - 2 \cdot 98$: $1 = 2 \cdot 98 - 39 \cdot (201 - 2 \cdot 98) = 80 \cdot 98 - 39 \cdot 201$.
5. Substitute $98$ from $299 = 201 \cdot 1 + 98 \Rightarrow 98 = 299 - 201$: $1 = 80 \cdot (299 - 201) - 39 \cdot 201 = 80 \cdot 299 - 119 \cdot 201$.

This shows that $-119$ is a solution for $x$. To express this as a positive integer modulo $299$, we compute $-119 + 299 = 180$. 

Thus, the multiplicative inverse of $201$ modulo $299$ is $180$.<|END_OF_LATENT|> The final answer is $180$.