Math Problem Statement

Find the number of integers between 1 and 10,000 inclusive which are divisible by 3 and 5, but not by either 7 or 11.

Solution

To solve the problem, we need to use principles of set theory and inclusion-exclusion. Let's denote:

  • n=10,000n = 10,000, the total range of integers.
  • AA: Integers divisible by 3.
  • BB: Integers divisible by 5.
  • CC: Integers divisible by 7.
  • DD: Integers divisible by 11.

We are tasked to find integers divisible by both 3 and 5 (i.e., 1515) but not divisible by 77 or 1111. Mathematically:

AB(CD)|A \cap B \setminus (C \cup D)|

Step 1: Total numbers divisible by 15

The numbers divisible by 15 are:

AB=n15=10,00015=666|A \cap B| = \left\lfloor \frac{n}{15} \right\rfloor = \left\lfloor \frac{10,000}{15} \right\rfloor = 666

Step 2: Numbers divisible by both 15 and 7

The least common multiple (LCM) of 1515 and 77 is 105105. The count is:

ABC=n105=10,000105=95|A \cap B \cap C| = \left\lfloor \frac{n}{105} \right\rfloor = \left\lfloor \frac{10,000}{105} \right\rfloor = 95

Step 3: Numbers divisible by both 15 and 11

The LCM of 1515 and 1111 is 165165. The count is:

ABD=n165=10,000165=60|A \cap B \cap D| = \left\lfloor \frac{n}{165} \right\rfloor = \left\lfloor \frac{10,000}{165} \right\rfloor = 60

Step 4: Numbers divisible by 15, 7, and 11

The LCM of 1515, 77, and 1111 is 11551155. The count is:

ABCD=n1155=10,0001155=8|A \cap B \cap C \cap D| = \left\lfloor \frac{n}{1155} \right\rfloor = \left\lfloor \frac{10,000}{1155} \right\rfloor = 8

Step 5: Inclusion-Exclusion to find AB(CD)|A \cap B \cap (C \cup D)|

The numbers divisible by 1515 and 77 or 1111 are:

AB(CD)=ABC+ABDABCD|A \cap B \cap (C \cup D)| = |A \cap B \cap C| + |A \cap B \cap D| - |A \cap B \cap C \cap D|

Substitute the values:

AB(CD)=95+608=147|A \cap B \cap (C \cup D)| = 95 + 60 - 8 = 147

Step 6: Subtract from total numbers divisible by 15

The count of integers divisible by 1515 but not 77 or 1111 is:

AB(CD)=ABAB(CD)|A \cap B \setminus (C \cup D)| = |A \cap B| - |A \cap B \cap (C \cup D)|

Substitute the values:

AB(CD)=666147=519|A \cap B \setminus (C \cup D)| = 666 - 147 = 519

Final Answer:

The number of integers between 1 and 10,000 inclusive that are divisible by 3 and 5 but not by 7 or 11 is:

519\boxed{519}

Let me know if you want a detailed breakdown or additional examples! 😊


  1. How does inclusion-exclusion work in set problems?
  2. Can we generalize this approach to other divisors?
  3. What is the importance of calculating LCM in such problems?
  4. How do you interpret modular arithmetic in divisibility rules?
  5. What if the range was extended to a different number, say 50,000?

Tip: For large ranges, writing modular rules as functions helps automate the calculations!

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Math Problem Analysis

Mathematical Concepts

Set Theory
Inclusion-Exclusion Principle
Divisibility
Least Common Multiple (LCM)

Formulas

|A ∩ B| = ⌊n / LCM(3, 5)⌋
|A ∩ B ∩ C| = ⌊n / LCM(15, 7)⌋
|A ∩ B ∩ D| = ⌊n / LCM(15, 11)⌋
|A ∩ B ∩ C ∩ D| = ⌊n / LCM(15, 7, 11)⌋
|A ∩ B \ (C ∪ D)| = |A ∩ B| - |A ∩ B ∩ (C ∪ D)|

Theorems

Inclusion-Exclusion Principle

Suitable Grade Level

Grades 9-12