Let $n$ be the smallest nonprime [integer](https://artofproblemsolving.com/wiki/index.php/Integer) greater than $1$ with no [prime factor](https://artofproblemsolving.com/wiki/index.php/Prime_factorization) less than $10$. Then 
$\mathrm{(A) \ }100<n\leq110 \qquad \mathrm{(B) \ }110<n\leq120 \qquad \mathrm{(C) \ } 120<n\leq130 \qquad \mathrm{(D) \ }130<n\leq140 \qquad \mathrm{(E) \ } 140<n\leq150$
Since the number isn't [prime](https://artofproblemsolving.com/wiki/index.php/Prime), it is a product of two primes. If the least integer were a product of more than two primes, then one prime could be removed without making the number prime or introducing any prime factors less than $10$. These prime factors must be greater than $10$, so the least prime factor is $11$. Therefore, the least integer is $11^2=121$, which is in $\boxed{\text{C}}$.