10. A sequence of positive integers starts with one 1 , followed by two $2 \mathrm{~s}$, three $3 \mathrm{~s}$, and so on. (Each positive integer $n$ occurs $n$ times.) How many of the first 105 numbers in this sequence are divisible by 3 ?
A 4
B 12
C 21
D 30
E 45
10. D The number of terms in the sequence up to and including the $n$ occurrences of $n$ is $1+2+3+\ldots+n=\frac{1}{2} n(n+1)$, which equals 105 when $n=14$. Hence the numbers divisible by 3 are 3 (three times), 6 (six times), 9 (nine times) and 12 (twelve times), giving a total of $3+6+9+12=30$ numbers.