problem solving

Question 1

Number of elements with odd factors in given range

Answer 1

Okay, the question “Number of elements with odd factors in a given range” is asking to count how many numbers within a specified range have an odd number of factors.

Understanding Numbers with an Odd Number of Factors:

A number has an odd number of factors if and only if it is a perfect square.

Here’s why:

Factors of a number typically come in pairs. For example, the factors of 12 are (1, 12), (2, 6), (3, 4). There are 6 factors (an even number).
The only case where a factor doesn’t have a distinct pair is when the factor is multiplied by itself to get the number (i.e., the number is a perfect square). For example, the factors of 16 are (1, 16), (2, 8), (4, 4). The factor 4 is paired with itself. If you count distinct factors (1, 2, 4, 8, 16), there are 5 factors (an odd number).
So, the problem reduces to finding the number of perfect squares within the given range [A, B] (inclusive).

Finding the Number of Perfect Squares in a Range [A, B]:

To count the number of perfect squares between A and B (inclusive), you need to find:

The smallest perfect square that is greater than or equal to A.
The largest perfect square that is less than or equal to B.
Let the range of numbers be from A to B. We are looking for integers k such that A ≤ k² ≤ B.

Taking the square root of all parts of the inequality:
√A ≤ k ≤ √B

Since k must be an integer, we are looking for integers k between ceil(√A) and floor(√B).

ceil(√A): The smallest integer whose square is greater than or equal to A.
floor(√B): The largest integer whose square is less than or equal to B.
Algorithm:

Given the range [A, B].
Calculate the integer part of the square root of B: upper_sqrt = floor(sqrt(B)). This gives you the largest integer k such that k² ≤ B.
Calculate the integer part of the square root of A-1: lower_sqrt = floor(sqrt(A - 1)). This gives you the largest integer k such that k² ≤ A-1.
The number of perfect squares in the range [A, B] is upper_sqrt - lower_sqrt.
Example: Find the number of elements with odd factors in the range [10, 50].

A = 10, B = 50.
sqrt(50) ≈ 7.07. upper_sqrt = floor(7.07) = 7. (The perfect squares less than or equal to 50 are 1², 2², 3², …, 7²).
A - 1 = 9. sqrt(9) = 3. lower_sqrt = floor(3) = 3. (The perfect squares less than or equal to 9 are 1², 2², 3²).
Number of perfect squares in [10, 50] = upper_sqrt - lower_sqrt = 7 - 3 = 4.
The perfect squares in the range [10, 50] are 16 (4²), 25 (5²), 36 (6²), and 49 (7²), which is indeed 4 numbers.

This method efficiently counts the perfect squares by finding the count of perfect squares up to B and subtracting the count of perfect squares up to A-1.