Math Problem Statement

n = 2001

Solution

Given N=2001N = 2001, let's calculate how many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5.

Step 1: Calculate the relevant multiples

  1. Multiples of 3 up to 2001: 20013=667=667\left\lfloor \frac{2001}{3} \right\rfloor = \left\lfloor 667 \right\rfloor = 667

  2. Multiples of 4 up to 2001: 20014=500.25=500\left\lfloor \frac{2001}{4} \right\rfloor = \left\lfloor 500.25 \right\rfloor = 500

  3. Multiples of 12 up to 2001 (common multiples of 3 and 4): 200112=166.75=166\left\lfloor \frac{2001}{12} \right\rfloor = \left\lfloor 166.75 \right\rfloor = 166

Using the inclusion-exclusion principle: Multiples of 3 or 4=667+500166=1001\text{Multiples of 3 or 4} = 667 + 500 - 166 = 1001

Step 2: Subtract the multiples of 5

  1. Multiples of 5 up to 2001: 20015=400.2=400\left\lfloor \frac{2001}{5} \right\rfloor = \left\lfloor 400.2 \right\rfloor = 400

  2. Multiples of 15 up to 2001 (multiples of both 3 and 5): 200115=133.4=133\left\lfloor \frac{2001}{15} \right\rfloor = \left\lfloor 133.4 \right\rfloor = 133

  3. Multiples of 20 up to 2001 (multiples of both 4 and 5): 200120=100.05=100\left\lfloor \frac{2001}{20} \right\rfloor = \left\lfloor 100.05 \right\rfloor = 100

  4. Multiples of 60 up to 2001 (multiples of 3, 4, and 5): 200160=33.35=33\left\lfloor \frac{2001}{60} \right\rfloor = \left\lfloor 33.35 \right\rfloor = 33

Now, subtracting the multiples of 5: Multiples of 3 or 4 but not 5=1001(133+10033)=1001200=801\text{Multiples of 3 or 4 but not 5} = 1001 - (133 + 100 - 33) = 1001 - 200 = 801

Final Answer:

There are 801 positive integers not exceeding 2001 that are multiples of 3 or 4 but not 5.

Would you like more details or have any questions?


Here are 5 related questions:

  1. How would the count change if we considered numbers up to 5000?
  2. What would be the count if we included multiples of 6 as well?
  3. How does the inclusion-exclusion principle help avoid over-counting?
  4. Can you calculate the number of multiples of 3 and 4 but not multiples of 2?
  5. What is the significance of finding common multiples, like those of 12, 15, and 20, in this problem?

Tip: Double-check your work by re-evaluating edge cases, such as considering the largest multiple within the given range. This helps ensure accuracy, especially when using the inclusion-exclusion principle.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Number Theory
Counting
Inclusion-Exclusion Principle

Formulas

-

Theorems

-

Suitable Grade Level

Grades 10-12