How many positive three-digit integers are divisible by both 11 and 5?
To be divisible by both 5 and 11, an integer must be a multiple of 55.  The smallest three-digit multiple of 55 is $2 \cdot 55 = 110,$ and the largest three-digit multiple of 55 is $18 \cdot 55 = 990$. So we can count the number of integers by the number of multiples, $2, 3, \ldots , 17 , 18$, of which there are $\boxed{17}$.
17
Level 3
Number Theory
Problem: How many positive three-digit integers are divisible by both 11 and 5?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of positive three-digit integers divisible by both 11 and 5, we first determine the least common multiple (LCM) of 11 and 5. Since both numbers are prime, their LCM is their product: LCM(11, 5) = 11 * 5 = 55.

Next, we identify the range of three-digit integers, which is from 100 to 999. We need to find the smallest three-digit integer that is divisible by 55. We do this by dividing 100 by 55 and rounding up to the nearest whole number. 

100 / 55 ≈ 1.818. Rounding up gives us 2. We then multiply 2 by 55 to find the smallest three-digit integer: 2 * 55 = 110.

Now, we find the largest three-digit integer divisible by 55. We divide 999 by 55 and round down to the nearest whole number.

999 / 55 ≈ 18.163. Rounding down gives us 18. We then multiply 18 by 55 to find the largest three-digit integer: 18 * 55 = 990.

Next, we determine how many multiples of 55 exist between 110 and 990. The multiples of 55 can be expressed as 55n, where n is an integer. 

The smallest n is 2 (for 110) and the largest n is 18 (for 990). To find the total count of integers n, we calculate the range: 

n = 18 - 2 + 1 = 17.

Thus, the total number of positive three-digit integers that are divisible by both 11 and 5 is 17.<|END_OF_LATENT|> The final answer is $17$.