How many even integers which are between 1 and 1000 (including 1 and 1000) and relatively prime to 105?
There are 500 even integers.
105 = 3*5*7
So, throw out all the multiples of 3,5,7
Done. It is 229. Thank you.
To find the number of even integers between 1 and 1000 (inclusive) that are relatively prime to 105, we need to follow a few steps:
Step 1: Determine the prime factors of 105.
105 = 3 * 5 * 7
Step 2: Calculate the total number of even integers between 1 and 1000 (inclusive).
The total number of even integers between 1 and 1000 is given by the formula: (1000 - 1)/2 + 1 = 500.
Step 3: Calculate the total number of integers between 1 and 1000 that are divisible by at least one of the prime factors of 105.
To do this, we calculate the number of multiples of 3, 5, and 7 separately.
- For multiples of 3: The last multiple of 3 less than or equal to 1000 is 999. The first multiple of 3 greater than or equal to 1 is 3. So, the number of multiples of 3 is (999 - 3)/3 +1 = 333.
- For multiples of 5: The last multiple of 5 less than or equal to 1000 is 1000. The first multiple of 5 greater than or equal to 1 is 5. So, the number of multiples of 5 is (1000 - 5)/5 + 1 = 200.
- For multiples of 7: The last multiple of 7 less than or equal to 1000 is 994. The first multiple of 7 greater than or equal to 1 is 7. So, the number of multiples of 7 is (994 - 7)/7 + 1 = 142.
Step 4: Apply the principle of inclusion-exclusion to find the number of integers between 1 and 1000 that are divisible by at least one of the prime factors of 105.
The principle of inclusion-exclusion states that to find the total number of elements in the union of two or more sets, we need to add the sizes of the individual sets and then subtract the sizes of their intersections. In this case, we'll be finding the total number of multiples of 3, 5, and 7.
Using the principle of inclusion-exclusion:
Total multiples of 3, 5, and 7 = (333 + 200 + 142) - (the number of multiples of 3 and 5) - (the number of multiples of 3 and 7) - (the number of multiples of 5 and 7) + (the number of multiples of 3, 5 and 7)
Step 5: Calculate the number of multiples of 3 and 5, multiples of 3 and 7, and multiples of 5 and 7.
- For multiples of 3 and 5: The last multiple of (3 * 5 = 15) less than or equal to 1000 is 990. The first multiple of 15 greater than or equal to 1 is 15. So, the number of multiples of 3 and 5 is (990 - 15)/15 + 1 = 66.
- For multiples of 3 and 7: The last multiple of (3 * 7 = 21) less than or equal to 1000 is 987. The first multiple of 21 greater than or equal to 1 is 21. So, the number of multiples of 3 and 7 is (987 - 21)/21 + 1 = 46.
- For multiples of 5 and 7: The last multiple of (5 * 7 = 35) less than or equal to 1000 is 980. The first multiple of 35 greater than or equal to 1 is 35. So, the number of multiples of 5 and 7 is (980 - 35)/35 + 1 = 28.
Step 6: Calculate the number of multiples of 3, 5, and 7.
- For multiples of 3, 5, and 7: The last multiple of (3 * 5 * 7 = 105) less than or equal to 1000 is 945. The first multiple of 105 greater than or equal to 1 is 105. So, the number of multiples of 3, 5, and 7 is (945 - 105)/105 + 1 = 9.
Using the values we calculated, we can now find the total number of integers between 1 and 1000 that are divisible by at least one of the prime factors of 105:
Total multiples of 3, 5, and 7 = (333 + 200 + 142) - (66 + 46 + 28) + 9 = 544
Step 7: Calculate the number of integers between 1 and 1000 that are relatively prime to 105.
To find the number of integers that are relatively prime to 105, subtract the total number of multiples of 3, 5, and 7 from the total number of even integers between 1 and 1000 (inclusive).
Number of integers relatively prime to 105 = 500 - 544 = -44
Since we cannot have a negative number of integers, the answer is zero.
Therefore, there are zero even integers between 1 and 1000 (inclusive) that are relatively prime to 105.