Find the smallest positive integer divisible by $10$, $11$, and $12$.
Factoring all three numbers, we find that $10=2\cdot 5$, $11=11$, and $12=2^2\cdot 3$. Taking the highest power of each, we see that the least common multiple of the three numbers is $2^2\cdot 3\cdot 5\cdot 11=60\cdot 11=\boxed{660}$.
660
Level 3
Prealgebra
Problem: Find the smallest positive integer divisible by $10$, $11$, and $12$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the smallest positive integer divisible by 10, 11, and 12, we need to determine the least common multiple (LCM) of these three numbers. 

First, we factor each number into its prime factors:
- 10 = 2 × 5
- 11 = 11 (11 is a prime number)
- 12 = 2^2 × 3

Next, we identify the highest power of each prime factor present in the factorizations:
- The prime factor 2 appears in 12 as 2^2.
- The prime factor 3 appears in 12 as 3^1.
- The prime factor 5 appears in 10 as 5^1.
- The prime factor 11 appears in 11 as 11^1.

Now, we take the highest powers of all prime factors:
- For 2, the highest power is 2^2.
- For 3, the highest power is 3^1.
- For 5, the highest power is 5^1.
- For 11, the highest power is 11^1.

The LCM is then calculated by multiplying these highest powers together:
LCM = 2^2 × 3^1 × 5^1 × 11^1.

Calculating this step-by-step:
1. 2^2 = 4.
2. 4 × 3 = 12.
3. 12 × 5 = 60.
4. 60 × 11 = 660.

Thus, the smallest positive integer that is divisible by 10, 11, and 12 is 660.<|END_OF_LATENT|> The final answer is $660$.