How Many Positive Integers Less Than 100 Are Divisible by Exactly One of 2, 5, or 7?

The question arises, how many positive integers less than 100 are divisible by exactly one of the numbers 2, 5, or 7? This exploration involves understanding the divisibility rules and prime factors, as well as using mathematical methods to find the solution.

The Unique Case and Its Explanation

The unique answer to this question is 70. It is the only positive integer less than 100 that is divisible by 2, 5, and 7. These are the only prime factors, and therefore, 70 is the product of these primes: 257 70. This makes 70 the only integer meeting the criteria of having non-trivial factors as a single 2, a single 5, and a single 7.

Understanding the Problem with Mathematical Framework

To tackle this problem, we use the concept of floor division to determine the count of numbers divisible by 2, 5, or 7. Specifically:

?99/2? 49 positive integers less than 100 are divisible by 2. ?99/5? 19 positive integers less than 100 are divisible by 5. ?99/7? 14 positive integers less than 100 are divisible by 7.

Next, we need to subtract the numbers that meet the criteria for two of these primes.

Subtracting Overlapping Multiples

For multiples of both 2 and 5 (10), we have:

?99/10? 9 numbers are divisible by 2 and 5.

For multiples of both 2 and 7 (14), we have:

?99/14? 7 numbers are divisible by 2 and 7.

For multiples of both 5 and 7 (35), we have:

?99/35? 2 numbers are divisible by 5 and 7.

Finally, for multiples that are divisible by all three (2, 5, and 7) excluding the unique case 70, we need to exclude these overlaps:

For multiples of 2, 5, and 7, we have 49 - 9 - 7 1 34 numbers that are only divisible by 2. For multiples of 2, 5, and 7, we have 19 - 9 - 2 1 9 numbers that are only divisible by 5. For multiples of 2, 5, and 7, we have 14 - 7 - 2 1 6 numbers that are only divisible by 7.

Adding these together:

34 9 6 49

Thus, there are 49 positive integers less than 100 which are divisible exactly by one of 2, 5, or 7.

Numerical Confirmation

Here's a Python script to numerically confirm the answer:

t  0
for n in range(1, 100):
    c2  n % 2  0
    c5  n % 5  0
    c7  n % 7  0
    t   (c2 or c5 or c7) and not (c2 and c5) and not (c2 and c7) and not (c5 and c7)
print(t)

This program confirms that there are indeed 49 such numbers.

By breaking down the problem and using the principles of floor division and exclusion, we arrive at the solution. Understanding the unique case of 70 and the overlaps for other multiples allows us to accurately count the integers that meet the criteria.