How many natural numbers lie strictly between the square root of 261 and the square root of 45109?

Introduction / Context:
This question checks your ability to work with square roots and count integers in a range described using inequalities. Instead of asking directly for numbers between two integers, the problem uses square roots of non perfect squares, which forces you to approximate or reason about their integer parts and then count natural numbers between them.

Given Data / Assumptions:

• We consider the square root of 261, written as sqrt(261).
• We also consider the square root of 45109, written as sqrt(45109).
• We must find how many natural numbers lie strictly between these two square roots.
• Natural numbers are positive integers 1, 2, 3, and so on.

Concept / Approach:
The key idea is that if a real number t satisfies a < t < b, then the integers between a and b are counted by looking at the smallest integer greater than a and the largest integer less than b. In this case, since 261 and 45109 are not perfect squares, their square roots are not integers. We approximate or bound them to determine the first and last integers inside the open interval (sqrt(261), sqrt(45109)). Then we count how many integers fall in that interval.

Step-by-Step Solution:
Step 1: Observe that 16^2 = 256 and 17^2 = 289. Step 2: Since 261 is between 256 and 289, we know that 16 < sqrt(261) < 17. Step 3: Next, note that 212^2 = 44944 and 213^2 = 45369. Step 4: Since 45109 lies between 44944 and 45369, we have 212 < sqrt(45109) < 213. Step 5: We are asked to find natural numbers strictly between sqrt(261) and sqrt(45109), so we need integers n such that sqrt(261) < n < sqrt(45109). Step 6: From the bounds, this means we need integers strictly greater than 16 and strictly less than 213, but we have a tighter upper bound of less than 213, and sqrt(45109) is already between 212 and 213. Step 7: Therefore the smallest integer greater than sqrt(261) is 17. Step 8: The largest integer less than sqrt(45109) is 212. Step 9: The natural numbers satisfying the condition are 17, 18, 19, up to 212. Step 10: The count of integers from 17 to 212 is 212 - 17 + 1 = 196.

Verification / Alternative check:
One way to check is simply to compute the number of integers from 1 to 212, which is 212, and subtract the number from 1 to 16, which is 16. The result 212 - 16 = 196 again gives the count of integers from 17 to 212 inclusive. Since we have already shown that these are exactly the integers between the two square roots, this confirms that the answer is correct. No other integers can lie in the interval because any integer less than 17 would be less than sqrt(261), and any integer greater than 212 would exceed sqrt(45109).

Why Other Options Are Wrong:
An answer of 144, 168, 195, or 200 would imply a shorter or longer range of integers. Such numbers come from incorrect bounds, such as using 18 instead of 17, or 211 instead of 212, or from miscounting the length of the interval. Precise identification of the lower and upper integer bounds is essential, and small off by one errors easily lead to those incorrect options.

Common Pitfalls:
A frequent mistake is forgetting that the interval is strict, so students sometimes include 16 or 213, which should not be counted. Another common issue is miscalculating the squares of 16, 17, 212, or 213, which shifts the bounding integers. Some learners also forget to add one when counting integers from m to n inclusive, leading to answers that are exactly one less than the true count. Careful work with inequalities and attentive counting help prevent these errors.

Final Answer:
The number of natural numbers between the two square roots is 196.