jmc

Let our answer be $n$. Write $n=42a+b$, where $a,b$ are positive integers and $0\le b<42$. Then note that $b,b+42,...,b+42(a-1)$ are all primes. If $b$ is $0\mathrm{mod}5$, then $b=5$ because $5$ is the only prime divisible by $5$. We get $n=215$ as our largest possibility in this case. If $b$ is $1\mathrm{mod}5$, then $b+2\times 42$ is divisible by $5$ and thus $a\le 2$. Thus, $n\le 3\times 42=126<215$. If $b$ is $2\mathrm{mod}5$, then $b+4\times 42$ is divisible by $5$ and thus $a\le 4$. Thus, $n\le 5\times 42=210<215$. If $b$ is $3\mathrm{mod}5$, then $b+1\times 42$ is divisible by $5$ and thus $a=1$. Thus, $n\le 2\times 42=84<215$. If $b$ is $4\mathrm{mod}5$, then $b+3\times 42$ is divisible by $5$ and thus $a\le 3$. Thus, $n\le 4\times 42=168<215$. Our answer is $\boxed{215}$.