The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is an arithmetic sequence with common difference 3 and $a_{1}=1$. The sequence $b_{1}, b_{2}, b_{3}, \ldots$ is an arithmetic sequence with common difference 10 and $b_{1}=2$. What is the smallest integer larger than 2023 that appears in both sequences?
Since $a_{1}, a_{2}, a_{3}, \ldots$ has $a_{1}=1$ and a common difference of 3 , we have that $a_{n}=1+3(n-1)$ for all $n \geq 1$. By similar reasoning, we get that $b_{m}=2+10(m-1)$ for all $m \geq 1$.

For fixed $m$, the term $b_{m}$ appears in both sequences exactly when there is some $n$ such that $a_{n}=b_{m}$.

Using the closed forms established earlier, the term $b_{m}$ appears in both sequences exactly when there exists $n$ such that $1+3(n-1)=2+10(m-1)$, which is equivalent to $n=\frac{10 m-6}{3}$.

In other words, $b_{m}$ is in both sequences exactly when the expression $\frac{10 m-6}{3}$ is a positive integer, which happens exactly when $10 m-6$ is a positive multiple of 3 .

Note that $m$ is a positive integer, so $10 m-6$ is at least 4 , which means $10 m-6$ is positive for all $m$.

Since 6 is a multiple of $3,10 \mathrm{~m}-6$ is a multiple of 3 exactly when $10 \mathrm{~m}$ is a multiple of 3 , but 3 and 10 have no prime factors in common, $10 \mathrm{~m}$ is a multiple of 3 exactly when $m$ is a multiple of 3 .

We have now shown that $b_{m}$ is in both sequences exactly when $m$ is a multiple of 3 .

Let $m=3 k$ for some $k \geq 1$. Then $b_{m}=2+10(3 k-1)=30 k-8$ and we wish to find the smallest possible $k$ for which $30 k-8>2023$.

This inequality is equivalent to $30 k>2031$ which is equivalent to $k>\frac{2031}{30}=67+\frac{21}{30}$. The smallest positive integer $k$ with this property is $k=68$. Therefore,

$$
b_{3 \times 68}=b_{204}=2+10(204-1)=2+10(203)=2032
$$

is the smallest integer that is larger than 2023 and is in both sequences.

ANSWER: 2032