We can solve this problem using the principle of inclusion and exclusion (PIE). Let's determine how many numbers between 1 and 250 are divisible by 2, 3, 5, or 7, and then subtract that from the total number of integers (250) to find those that are not divisible by any of these numbers.

Step 1: Define Sets

Let:

• $A_2$: numbers divisible by 2
• $A_3$: numbers divisible by 3
• $A_5$: numbers divisible by 5
• $A_7$: numbers divisible by 7

We want to calculate the size of the union of these sets, which is the number of integers divisible by at least one of 2, 3, 5, or 7.

Step 2: Use Inclusion-Exclusion

The principle of inclusion-exclusion tells us: $|A_2 \cup A_3 \cup A_5 \cup A_7| = |A_2| + |A_3| + |A_5| + |A_7| - |A_2 \cap A_3| - |A_2 \cap A_5| - \cdots + |A_2 \cap A_3 \cap A_5| + \cdots - |A_2 \cap A_3 \cap A_5 \cap A_7|$

Step 3: Compute Individual Terms

• $|A_2| = \left\lfloor \frac{250}{2} \right\rfloor = 125$
• $|A_3| = \left\lfloor \frac{250}{3} \right\rfloor = 83$
• $|A_5| = \left\lfloor \frac{250}{5} \right\rfloor = 50$
• $|A_7| = \left\lfloor \frac{250}{7} \right\rfloor = 35$

Step 4: Compute Pairwise Intersections

• $|A_2 \cap A_3| = \left\lfloor \frac{250}{6} \right\rfloor = 41$ (LCM of 2 and 3 is 6)
• $|A_2 \cap A_5| = \left\lfloor \frac{250}{10} \right\rfloor = 25$ (LCM of 2 and 5 is 10)
• $|A_2 \cap A_7| = \left\lfloor \frac{250}{14} \right\rfloor = 17$ (LCM of 2 and 7 is 14)
• $|A_3 \cap A_5| = \left\lfloor \frac{250}{15} \right\rfloor = 16$ (LCM of 3 and 5 is 15)
• $|A_3 \cap A_7| = \left\lfloor \frac{250}{21} \right\rfloor = 11$ (LCM of 3 and 7 is 21)
• $|A_5 \cap A_7| = \left\lfloor \frac{250}{35} \right\rfloor = 7$ (LCM of 5 and 7 is 35)

Step 5: Compute Triple Intersections

• $|A_2 \cap A_3 \cap A_5| = \left\lfloor \frac{250}{30} \right\rfloor = 8$ (LCM of 2, 3, and 5 is 30)
• $|A_2 \cap A_3 \cap A_7| = \left\lfloor \frac{250}{42} \right\rfloor = 5$ (LCM of 2, 3, and 7 is 42)
• $|A_2 \cap A_5 \cap A_7| = \left\lfloor \frac{250}{70} \right\rfloor = 3$ (LCM of 2, 5, and 7 is 70)
• $|A_3 \cap A_5 \cap A_7| = \left\lfloor \frac{250}{105} \right\rfloor = 2$ (LCM of 3, 5, and 7 is 105)

Step 6: Compute Four-Way Intersection

• $|A_2 \cap A_3 \cap A_5 \cap A_7| = \left\lfloor \frac{250}{210} \right\rfloor = 1$ (LCM of 2, 3, 5, and 7 is 210)

Step 7: Apply Inclusion-Exclusion Formula

$|A_2 \cup A_3 \cup A_5 \cup A_7| = 125 + 83 + 50 + 35 - (41 + 25 + 17 + 16 + 11 + 7) + (8 + 5 + 3 + 2) - 1$ $= 293 - 117 + 18 - 1 = 193$

Step 8: Final Answer

The number of integers between 1 and 250 that are divisible by 2, 3, 5, or 7 is 193. Therefore, the number of integers not divisible by 2, 3, 5, or 7 is: $250 - 193 = 57$

So, there are 57 integers between 1 and 250 that are not divisible by any of the numbers 2, 3, 5, or 7.

Would you like more details on any of the steps?

Related Questions:

1. How can the principle of inclusion-exclusion be applied to different sets?
2. How many numbers between 1 and 1000 are divisible by 2, 3, or 5?
3. How many prime numbers are there between 1 and 250?
4. What are the multiples of 2, 3, 5, and 7 between 1 and 100?
5. How would the result change if we added divisibility by 11 as well?

Tip:

Inclusion-Exclusion is a powerful technique that can handle complex overlap problems by systematically adding and subtracting overlaps between sets.