| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the number of integers between 1 and 100 (inclusive) that are divisible by either 2, 3, or 5, we can use the principle of inclusion-exclusion. First, we find the number of integers divisible by each of the numbers 2, 3, and 5: - Divisible by 2: 100/2 = 50 integers - Divisible by 3: 100/3 = 33 integers (rounding down to 33) - Divisible by 5: 100/5 = 20 integers Next, we find the number of integers divisible by the combinations of these numbers (i.e., divisible by 2 and 3, 2 and 5, 3 and 5): - Divisible by 2 and 3 (i.e., by 6): 100/6 = 16 integers (rounding down to 16) - Divisible by 2 and 5 (i.e., by 10): 100/10 = 10 integers - Divisible by 3 and 5 (i.e., by 15): 100/15 = 6 integers (rounding down to 6) Lastly, we find the number of integers divisible by all three numbers (2, 3, and 5): - Divisible by 2, 3, and 5 (i.e., by 30): 100/30 = 3 integers (rounding down to 3) Now, we use the principle of inclusion-exclusion to find the total number of integers divisible by 2, 3, or 5: Total = Divisible by 2 + Divisible by 3 + Divisible by 5 - Divisible by 2 and 3 - Divisible by 2 and 5 - Divisible by 3 and 5 + Divisible by 2, 3, and 5 Total = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 Therefore, there are 74 integers between 1 and 100 (inclusive) that are divisible by 2, 3, or 5. |
74 |
| deepseekr1 | The number of integers between 1 and 100 divisible by 2, 3, or 5 can be found using the principle of inclusion-exclusion: 1. **Count individual sets**: - Divisible by 2: \( \left\lfloor \frac{100}{2} \right\rfloor = 50 \) - Divisible by 3: \( \left\lfloor \frac{100}{3} \right\rfloor = 33 \) - Divisible by 5: \( \left\lfloor \frac{100}{5} \right\rfloor = 20 \) 2. **Subtract pairwise intersections**: - LCM(2,3)=6: \( \left\lfloor \frac{100}{6} \right\rfloor = 16 \) - LCM(2,5)=10: \( \left\lfloor \frac{100}{10} \right\rfloor = 10 \) - LCM(3,5)=15: \( \left\lfloor \frac{100}{15} \right\rfloor = 6 \) 3. **Add back the triple intersection**: - LCM(2,3,5)=30: \( \left\lfloor \frac{100}{30} \right\rfloor = 3 \) Applying inclusion-exclusion: \[ 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 \] **Answer**: \boxed{74} |
74 |
| gpto1 | To solve this problem, we need to find the number of integers between 1 and 100 (inclusive) that are divisible by 2, 3, or 5. We'll use the principle of inclusion-exclusion to avoid counting numbers multiple times. **Step 1: Find the counts of multiples of each number individually.** - **Multiples of 2:** \[ \left\lfloor \frac{100}{2} \right\rfloor = 50 \] - **Multiples of 3:** \[ \left\lfloor \frac{100}{3} \right\rfloor = 33 \] - **Multiples of 5:** \[ \left\lfloor \frac{100}{5} \right\rfloor = 20 \] **Step 2: Subtract the counts of numbers that are multiples of two numbers (since they've been counted twice).** - **Multiples of both 2 and 3 (i.e., multiples of 6):** \[ \left\lfloor \frac{100}{6} \right\rfloor = 16 \] - **Multiples of both 2 and 5 (i.e., multiples of 10):** \[ \left\lfloor \frac{100}{10} \right\rfloor = 10 \] - **Multiples of both 3 and 5 (i.e., multiples of 15):** \[ \left\lfloor \frac{100}{15} \right\rfloor = 6 \] **Step 3: Add back the count of numbers that are multiples of all three (since they've been subtracted too many times).** - **Multiples of 2, 3, and 5 (i.e., multiples of 30):** \[ \left\lfloor \frac{100}{30} \right\rfloor = 3 \] **Step 4: Apply the inclusion-exclusion principle.** Total numbers divisible by 2, 3, or 5: \[ \text{Total} = (50 + 33 + 20) - (16 + 10 + 6) + 3 \] \[ \text{Total} = 103 - 32 + 3 \] \[ \text{Total} = 74 \] **Answer:** 74 |
74 |
| GPT-4o | To solve the problem of finding how many integers between 1 and 100 (inclusive) are divisible by either 2, 3, or 5, we can use the principle of Inclusion-Exclusion. Let's define: - \( A \) as the set of integers between 1 and 100 that are divisible by 2. - \( B \) as the set of integers between 1 and 100 that are divisible by 3. - \( C \) as the set of integers between 1 and 100 that are divisible by 5. We need to find the size of the union of these three sets, i.e., \( |A \cup B \cup C| \). According to the principle of Inclusion-Exclusion: \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \] First, we calculate the size of each individual set: - \( |A| \) is the number of integers between 1 and 100 divisible by 2. These integers form an arithmetic sequence: 2, 4, 6, ..., 100. The number of terms is: \( \frac{100 - 2}{2} + 1 = 50 \). - \( |B| \) is the number of integers between 1 and 100 divisible by 3. These integers form an arithmetic sequence: 3, 6, 9, ..., 99. The number of terms is: \( \frac{99 - 3}{3} + 1 = 33 \). - \( |C| \) is the number of integers between 1 and 100 divisible by 5. These integers form an arithmetic sequence: 5, 10, 15, ..., 100. The number of terms is: \( \frac{100 - 5}{5} + 1 = 20 \). Next, we calculate the size of the intersections of every pair of sets: - \( |A \cap B| \) is the number of integers between 1 and 100 divisible by both 2 and 3 (i.e., divisible by 6). These integers form an arithmetic sequence: 6, 12, 18, ..., 96. The number of terms is: \( \frac{96 - 6}{6} + 1 = 16 \). - \( |A \cap C| \) is the number of integers between 1 and 100 divisible by both 2 and 5 (i.e., divisible by 10). These integers form an arithmetic sequence: 10, 20, 30, ..., 100. The number of terms is: \( \frac{100 - 10}{10} + 1 = 10 \). - \( |B \cap C| \) is the number of integers between 1 and 100 divisible by both 3 and 5 (i.e., divisible by 15). These integers form an arithmetic sequence: 15, 30, 45, ..., 90. The number of terms is: \( \frac{90 - 15}{15} + 1 = 6 \). Finally, we calculate the size of the intersection of all three sets: - \( |A \cap B \cap C| \) is the number of integers between 1 and 100 divisible by 2, 3, and 5 (i.e., divisible by 30). These integers form an arithmetic sequence: 30, 60, 90. The number of terms is: \( \frac{90 - 30}{30} + 1 = 3 \). Putting it all together: \[ |A \cup B \cup C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 \] Therefore, the number of integers between 1 and 100 that are divisible by either 2, 3, or 5 is \( 74 \). |
74 |
| llama3.1-70B-Instruct | To solve this problem, we can use the principle of inclusion-exclusion. Let's first count the number of integers between 1 and 100 that are divisible by each of the numbers 2, 3, and 5: * Divisible by 2: 50 (since 2, 4, 6, ..., 100 are all divisible by 2) * Divisible by 3: 33 (since 3, 6, 9, ..., 99 are all divisible by 3) * Divisible by 5: 20 (since 5, 10, 15, ..., 100 are all divisible by 5) However, we have counted some numbers multiple times. For example, the number 30 is divisible by both 2 and 3, and the number 60 is divisible by all three numbers. To correct for this, we need to subtract the number of integers that are divisible by two or more of the numbers 2, 3, and 5. * Divisible by both 2 and 3: 16 (since 6, 12, 18, ..., 96 are all divisible by both 2 and 3) * Divisible by both 2 and 5: 10 (since 10, 20, 30, ..., 100 are all divisible by both 2 and 5) * Divisible by both 3 and 5: 6 (since 15, 30, 45, ..., 90 are all divisible by both 3 and 5) * Divisible by all three numbers: 3 (since 30, 60, and 90 are all divisible by 2, 3, and 5) Now, we can use the principle of inclusion-exclusion to find the total number of integers between 1 and 100 that are divisible by either 2, 3, or 5: Total = (Divisible by 2) + (Divisible by 3) + (Divisible by 5) - (Divisible by both 2 and 3) - (Divisible by both 2 and 5) - (Divisible by both 3 and 5) + (Divisible by all three numbers) = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 Therefore, there are 74 integers between 1 and 100 that are divisible by either 2, 3, or 5. |
74 |
| llama3.1-405B-Instruct | To find the number of integers between 1 and 100 that are divisible by either 2, 3, or 5, we can use the principle of inclusion-exclusion. 1. First, let's find the number of integers divisible by 2, 3, and 5 individually: - Divisible by 2: 100/2 = 50 - Divisible by 3: 100/3 = 33 (rounded down) - Divisible by 5: 100/5 = 20 2. Next, we need to find the number of integers divisible by the combinations of 2, 3, and 5: - Divisible by 2 and 3 (6): 100/6 = 16 (rounded down) - Divisible by 2 and 5 (10): 100/10 = 10 - Divisible by 3 and 5 (15): 100/15 = 6 (rounded down) - Divisible by 2, 3, and 5 (30): 100/30 = 3 (rounded down) 3. Now, we apply the principle of inclusion-exclusion: - Add the numbers divisible by 2, 3, and 5 individually: 50 + 33 + 20 = 103 - Subtract the numbers divisible by the combinations of 2, 3, and 5: 103 - (16 + 10 + 6) = 103 - 32 = 71 - Add back the number divisible by 2, 3, and 5: 71 + 3 = 74 Therefore, there are 74 integers between 1 and 100 that are divisible by either 2, 3, or 5. |
74 |
| DeepSeek-Coder-V2 | To solve the problem of finding how many integers between 1 and 100 are divisible by either 2, 3, or 5, we can use the principle of inclusion-exclusion. This principle helps us count the number of elements in the union of multiple sets by including and excluding their intersections appropriately. Let's define the sets: - \( A \) as the set of numbers divisible by 2. - \( B \) as the set of numbers divisible by 3. - \( C \) as the set of numbers divisible by 5. We need to find the size of the set \( |A \cup B \cup C| \). Using the principle of inclusion-exclusion for three sets, we have: \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \] First, we calculate the sizes of the individual sets: - \( |A| \) (numbers divisible by 2): There are 50 numbers between 1 and 100 that are divisible by 2. - \( |B| \) (numbers divisible by 3): There are 33 numbers between 1 and 100 that are divisible by 3. - \( |C| \) (numbers divisible by 5): There are 20 numbers between 1 and 100 that are divisible by 5. Next, we calculate the sizes of the intersections: - \( |A \cap B| \) (numbers divisible by both 2 and 3, i.e., by 6): There are 16 numbers between 1 and 100 that are divisible by 6. - \( |A \cap C| \) (numbers divisible by both 2 and 5, i.e., by 10): There are 10 numbers between 1 and 100 that are divisible by 10. - \( |B \cap C| \) (numbers divisible by both 3 and 5, i.e., by 15): There are 6 numbers between 1 and 100 that are divisible by 15. Finally, we calculate the size of the intersection of all three sets: - \( |A \cap B \cap C| \) (numbers divisible by 2, 3, and 5, i.e., by 30): There are 3 numbers between 1 and 100 that are divisible by 30. Now, we apply the principle of inclusion-exclusion: \[ |A \cup B \cup C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 \] Thus, the number of integers between 1 and 100 that are divisible by either 2, 3, or 5 is: \[ \boxed{74} \] |
74 |
| Qwen2-72B-Instruct | To find the number of integers between 1 and 100 (inclusive) that are divisible by 2, 3, or 5, we can use the principle of inclusion-exclusion. First, we find the number of integers divisible by each of the numbers 2, 3, and 5: - Divisible by 2: 100/2 = 50 integers - Divisible by 3: 100/3 = 33 integers (rounding down to 33) - Divisible by 5: 100/5 = 20 integers Next, we find the number of integers divisible by the combinations of these numbers (i.e., divisible by 2 and 3, 2 and 5, 3 and 5): - Divisible by 2 and 3 (i.e., by 6): 100/6 = 16 integers (rounding down to 16) - Divisible by 2 and 5 (i.e., by 10): 100/10 = 10 integers - Divisible by 3 and 5 (i.e., by 15): 100/15 = 6 integers (rounding down to 6) Lastly, we find the number of integers divisible by all three numbers (2, 3, and 5): - Divisible by 2, 3, and 5 (i.e., by 30): 100/30 = 3 integers (rounding down to 3) Now, we use the principle of inclusion-exclusion to find the total number of integers divisible by 2, 3, or 5: Total = Divisible by 2 + Divisible by 3 + Divisible by 5 - Divisible by 2 and 3 - Divisible by 2 and 5 - Divisible by 3 and 5 + Divisible by 2, 3, and 5 Total = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 Therefore, there are 74 integers between 1 and 100 (inclusive) that are divisible by 2, 3, or 5. |
74 |
| baichuan4 | To find the number of integers between 1 and 100 (inclusive) that are divisible by 2, 3, or 5, we can use the principle of inclusion-exclusion. First, we count how many numbers are divisible by each of the three divisors separately: - Divisible by 2: $\left\lfloor \frac{100}{2} \right\rfloor = 50$ - Divisible by 3: $\left\lfloor \frac{100}{3} \right\rfloor = 33$ - Divisible by 5: $\left\lfloor \frac{100}{5} \right\rfloor = 20$ Next, we subtract the numbers that are divisible by the combinations of these divisors (since they were counted more than once): - Divisible by 2 and 3: $\left\lfloor \frac{100}{2*3} \right\rfloor = \left\lfloor \frac{100}{6} \right\rfloor = 16$ - Divisible by 2 and 5: $\left\lfloor \frac{100}{2*5} \right\rfloor = \left\lfloor \frac{100}{10} \right\rfloor = 10$ - Divisible by 3 and 5: $\left\lfloor \frac{100}{3*5} \right\rfloor = \left\lfloor \frac{100}{15} \right\rfloor = 6$ Then, we add back the numbers that are divisible by all three divisors (since we subtracted them three times but they should only be subtracted once): - Divisible by 2, 3, and 5: $\left\lfloor \frac{100}{2*3*5} \right\rfloor = \left\lfloor \frac{100}{30} \right\rfloor = 3$ Now, we can calculate the total number of integers divisible by 2, 3, or 5 using the inclusion-exclusion principle: Total = (Divisible by 2) + (Divisible by 3) + (Divisible by 5) - (Divisible by 2 and 3) - (Divisible by 2 and 5) - (Divisible by 3 and 5) + (Divisible by 2, 3, and 5) Total = $50 + 33 + 20 - 16 - 10 - 6 + 3 = 74$ So, there are 74 integers between 1 and 100 (inclusive) that are divisible by either 2, 3, or 5. |
74 |
| doubao | The number of integers divisible by 2 between 1 and 100 is 100/2 = 50. The number of integers divisible by 3 between 1 and 100 is ⌊100/3⌋ = 33 (where ⌊ ⌋ represents the floor function). The number of integers divisible by 5 between 1 and 100 is 100/5 = 20. However, we have counted some numbers twice when they are divisible by both 2 and 3 (i.e., divisible by 6), both 2 and 5 (i.e., divisible by 10), and both 3 and 5 (i.e., divisible by 15). The number of integers divisible by 6 between 1 and 100 is ⌊100/6⌋ = 16. The number of integers divisible by 10 between 1 and 100 is ⌊100/10⌋ = 10. The number of integers divisible by 15 between 1 and 100 is ⌊100/15⌋ = 6. To find the total number of integers divisible by either 2, 3, or 5, we add up the individual counts and subtract the double-counted numbers: 50 + 33 + 20 - 16 - 10 - 6 = 71. So, there are 71 integers between 1 and 100 that are divisible by either 2, 3, or 5. |
74 |
| minimax | To solve this problem, we can use the principle of inclusion-exclusion. We need to count the number of integers divisible by 2, 3, and 5, but we must subtract the numbers that are divisible by the combinations of these numbers (like 2 and 3, 2 and 5, and 3 and 5) to avoid double-counting, and then add back the numbers that are divisible by all three (2, 3, and 5) since they have been subtracted three times. First, let's count the numbers divisible by each of the numbers: - Divisible by 2: There are 100/2 = 50 numbers between 1 and 100 that are divisible by 2. - Divisible by 3: There are 100/3 = 33 numbers between 1 and 100 that are divisible by 3. - Divisible by 5: There are 100/5 = 20 numbers between 1 and 100 that are divisible by 5. Next, we count the numbers divisible by the combinations of two of the numbers: - Divisible by 2 and 3 (6): There are 100/6 = 16 numbers between 1 and 100 that are divisible by both 2 and 3. - Divisible by 2 and 5 (10): There are 100/10 = 10 numbers between 1 and 100 that are divisible by both 2 and 5. - Divisible by 3 and 5 (15): There are 100/15 = 6 numbers between 1 and 100 that are divisible by both 3 and 5. Finally, we count the numbers divisible by all three: - Divisible by 2, 3, and 5 (30): There are 100/30 = 3 numbers between 1 and 100 that are divisible by 2, 3, and 5. Now we apply the principle of inclusion-exclusion: Total = (Divisible by 2) + (Divisible by 3) + (Divisible by 5) - (Divisible by 2 and 3) - (Divisible by 2 and 5) - (Divisible by 3 and 5) + (Divisible by 2, 3, and 5) Total = 50 + 33 + 20 - 16 - 10 - 6 + 3 Total = 103 - 32 + 3 Total = 71 + 3 Total = 74 Therefore, there are 74 integers between 1 and 100 that are divisible by either 2, 3, or 5. |
74 |
| kimi | To solve this problem, we can use the principle of Inclusion-Exclusion to count the integers that are divisible by 2, 3, or 5. Let's define: - \( A \) as the set of numbers divisible by 2. - \( B \) as the set of numbers divisible by 3. - \( C \) as the set of numbers divisible by 5. We want to find the number of elements in the union of these sets, \( |A \cup B \cup C| \). According to the principle of Inclusion-Exclusion: \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \] First, let's calculate each part: 1. \( |A| \): The number of multiples of 2 between 1 and 100 is \( \lfloor \frac{100}{2} \rfloor = 50 \). 2. \( |B| \): The number of multiples of 3 between 1 and 100 is \( \lfloor \frac{100}{3} \rfloor = 33 \). 3. \( |C| \): The number of multiples of 5 between 1 and 100 is \( \lfloor \frac{100}{5} \rfloor = 20 \). Next, we calculate the intersections: 4. \( |A \cap B| \): The number of multiples of both 2 and 3 (which is 6) between 1 and 100 is \( \lfloor \frac{100}{6} \rfloor = 16 \). 5. \( |A \cap C| \): The number of multiples of both 2 and 5 (which is 10) between 1 and 100 is \( \lfloor \frac{100}{10} \rfloor = 10 \). 6. \( |B \cap C| \): The number of multiples of both 3 and 5 (which is 15) between 1 and 100 is \( \lfloor \frac{100}{15} \rfloor = 6 \). Finally, we calculate the triple intersection: 7. \( |A \cap B \cap C| \): The number of multiples of 2, 3, and 5 (which is 30) between 1 and 100 is \( \lfloor \frac{100}{30} \rfloor = 3 \). Now we can plug these values into the formula: \[ |A \cup B \cup C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 \] So, there are 74 integers between 1 and 100 that are divisible by 2, 3, or 5. |
74 |
| wenxinyiyan | To solve this problem, we can use the principle of inclusion-exclusion. 1. **Count numbers divisible by 2**: * There are 100 numbers in total. * Half of them are divisible by 2, so there are 100/2 = 50 numbers divisible by 2. 2. **Count numbers divisible by 3**: * Every third number is divisible by 3. * So, there are 100/3 = 33 numbers divisible by 3 (we take the integer part). 3. **Count numbers divisible by 5**: * Every fifth number is divisible by 5. * So, there are 100/5 = 20 numbers divisible by 5. 4. **Count numbers divisible by 2 and 3 (i.e., divisible by 6)**: * Every sixth number is divisible by 6. * So, there are 100/6 = 16 numbers divisible by 6 (we take the integer part). 5. **Count numbers divisible by 2 and 5 (i.e., divisible by 10)**: * Every tenth number is divisible by 10. * So, there are 100/10 = 10 numbers divisible by 10. 6. **Count numbers divisible by 3 and 5 (i.e., divisible by 15)**: * Every fifteenth number is divisible by 15. * So, there are 100/15 = 6 numbers divisible by 15 (we take the integer part). 7. **Count numbers divisible by 2, 3, and 5 (i.e., divisible by 30)**: * Every thirtieth number is divisible by 30. * So, there are 100/30 = 3 numbers divisible by 30 (we take the integer part). Now, using the principle of inclusion-exclusion, we add the numbers divisible by 2, 3, or 5, and then subtract the numbers divisible by two of these (since they were counted twice), and finally add back the numbers divisible by all three (since they were subtracted twice and should only be counted once). So, the total number of integers between 1 and 100 (inclusive) that are divisible by either 2, 3, or 5 is: 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 Therefore, there are 74 integers between 1 and 100 (inclusive) that are divisible by either 2, 3, or 5. |
74 |
| zhipu | To solve this problem, we can use the principle of inclusion-exclusion. This principle allows us to find the size of the union of overlapping sets by considering the sizes of the individual sets, the sizes of their intersections, and then adjusting for any over-counting. Let's define the sets as follows: - Set A: the set of integers between 1 and 100 that are divisible by 2. - Set B: the set of integers between 1 and 100 that are divisible by 3. - Set C: the set of integers between 1 and 100 that are divisible by 5. We want to find the number of integers in the union of these sets, which we'll denote as |A ∪ B ∪ C|. According to the principle of inclusion-exclusion, we have: |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| First, we calculate the sizes of the individual sets: - |A| is the number of integers between 1 and 100 that are divisible by 2. Since every second integer is divisible by 2, |A| = 100 / 2 = 50. - |B| is the number of integers between 1 and 100 that are divisible by 3. Since every third integer is divisible by 3, |B| = 100 / 3 = 33 (we use the floor function because we're counting integers). - |C| is the number of integers between 1 and 100 that are divisible by 5. Since every fifth integer is divisible by 5, |C| = 100 / 5 = 20. Next, we calculate the sizes of the intersections of the sets: - |A ∩ B| is the number of integers between 1 and 100 that are divisible by both 2 and 3, which means they are divisible by the least common multiple (LCM) of 2 and 3, which is 6. So, |A ∩ B| = 100 / 6 = 16. - |A ∩ C| is the number of integers between 1 and 100 that are divisible by both 2 and 5, which means they are divisible by the LCM of 2 and 5, which is 10. So, |A ∩ C| = 100 / 10 = 10. - |B ∩ C| is the number of integers between 1 and 100 that are divisible by both 3 and 5, which means they are divisible by the LCM of 3 and 5, which is 15. So, |B ∩ C| = 100 / 15 = 6. Finally, we calculate the size of the intersection of all three sets: - |A ∩ B ∩ C| is the number of integers between 1 and 100 that are divisible by 2, 3, and 5, which means they are divisible by the LCM of 2, 3, and 5, which is 30. So, |A ∩ B ∩ C| = 100 / 30 = 3. Now we can plug these values into our inclusion-exclusion formula: |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| |A ∪ B ∪ C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 |A ∪ B ∪ C| = 74 So, there are 74 integers between 1 and 100 that are divisible by either 2, 3, or 5. |
74 |
| gpto1 | To solve this problem, we need to find the number of integers between 1 and 100 (inclusive) that are divisible by 2, 3, or 5. We'll use the principle of inclusion-exclusion to avoid counting numbers multiple times. **Step 1: Find the counts of multiples of each number individually.** - **Multiples of 2:** \[ \left\lfloor \frac{100}{2} \right\rfloor = 50 \] - **Multiples of 3:** \[ \left\lfloor \frac{100}{3} \right\rfloor = 33 \] - **Multiples of 5:** \[ \left\lfloor \frac{100}{5} \right\rfloor = 20 \] **Step 2: Subtract the counts of numbers that are multiples of two numbers (since they've been counted twice).** - **Multiples of both 2 and 3 (i.e., multiples of 6):** \[ \left\lfloor \frac{100}{6} \right\rfloor = 16 \] - **Multiples of both 2 and 5 (i.e., multiples of 10):** \[ \left\lfloor \frac{100}{10} \right\rfloor = 10 \] - **Multiples of both 3 and 5 (i.e., multiples of 15):** \[ \left\lfloor \frac{100}{15} \right\rfloor = 6 \] **Step 3: Add back the count of numbers that are multiples of all three (since they've been subtracted too many times).** - **Multiples of 2, 3, and 5 (i.e., multiples of 30):** \[ \left\lfloor \frac{100}{30} \right\rfloor = 3 \] **Step 4: Apply the inclusion-exclusion principle.** Total numbers divisible by 2, 3, or 5: \[ \text{Total} = (50 + 33 + 20) - (16 + 10 + 6) + 3 \] \[ \text{Total} = 103 - 32 + 3 \] \[ \text{Total} = 74 \] **Answer:** 74 |
74 |
| deepseekv3 | To determine how many integers between 1 and 100 (inclusive) are divisible by either 2, 3, or 5, we can use the **principle of inclusion-exclusion**. Here's the step-by-step solution: --- ### Step 1: Count the numbers divisible by 2, 3, or 5 individually 1. **Divisible by 2**: The numbers divisible by 2 are \( 2, 4, 6, \dots, 100 \). Total numbers = \( \frac{100}{2} = 50 \). 2. **Divisible by 3**: The numbers divisible by 3 are \( 3, 6, 9, \dots, 99 \). Total numbers = \( \frac{99}{3} = 33 \). 3. **Divisible by 5**: The numbers divisible by 5 are \( 5, 10, 15, \dots, 100 \). Total numbers = \( \frac{100}{5} = 20 \). --- ### Step 2: Subtract the overlaps (numbers divisible by two of the numbers) 1. **Divisible by both 2 and 3 (i.e., divisible by 6)**: The numbers divisible by 6 are \( 6, 12, 18, \dots, 96 \). Total numbers = \( \frac{96}{6} = 16 \). 2. **Divisible by both 2 and 5 (i.e., divisible by 10)**: The numbers divisible by 10 are \( 10, 20, 30, \dots, 100 \). Total numbers = \( \frac{100}{10} = 10 \). 3. **Divisible by both 3 and 5 (i.e., divisible by 15)**: The numbers divisible by 15 are \( 15, 30, 45, \dots, 90 \). Total numbers = \( \frac{90}{15} = 6 \). --- ### Step 3: Add back the overlaps (numbers divisible by all three numbers) 1. **Divisible by 2, 3, and 5 (i.e., divisible by 30)**: The numbers divisible by 30 are \( 30, 60, 90 \). Total numbers = \( \frac{90}{30} = 3 \). --- ### Step 4: Apply the inclusion-exclusion principle The total number of integers divisible by either 2, 3, or 5 is: \[ \text{Total} = (\text{Divisible by 2}) + (\text{Divisible by 3}) + (\text{Divisible by 5}) \\ - (\text{Divisible by 2 and 3}) - (\text{Divisible by 2 and 5}) - (\text{Divisible by 3 and 5}) \\ + (\text{Divisible by 2, 3, and 5}) \] Substitute the values: \[ \text{Total} = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74 \] --- ### Final Answer: **74** integers between 1 and 100 are divisible by either 2, 3, or 5. |
74 |