Numbers and Divisibility

The GCD of 18 and 24 is the largest positive integer that divides both numbers without leaving a remainder. By finding the prime factorization of each number, we have 18 = 2 * 3^2 and 24 = 2^3 * 3. The GCD is the product of the common prime factors raised to their lowest powers, which is 2 * 3 = 6.

To find the smallest positive integer divisible by 2, 3, and 5, we need to find their least common multiple (LCM). The LCM of 2, 3, and 5 is 30, which is the smallest number that is divisible by all three numbers.

To find the smallest positive integer n such that n! is divisible by 100, we need to find the smallest n for which n! contains at least two factors of 5 and two factors of 2. The smallest number that satisfies this condition is 25, as 25! contains 6 factors of 5 and 12 factors of 2.

To find the greatest integer n such that 2^n divides 1000!, we need to determine the highest power of 2 that divides 1000!. We can do this by repeatedly dividing 1000! by 2 until the result is no longer divisible by 2. After performing this process, we find that 2^26 divides 1000!, but 2^27 does not. Therefore, the greatest integer n is 26.

The elements of (S) are the multiples of 3 and 5 less than 100. The multiples of 3 are (3, 6, 9, ..., 99), and the multiples of 5 are (5, 10, 15, ..., 95). However, some numbers are counted twice, such as 15, which is a multiple of both 3 and 5. To find the sum of all the elements of (S), we need to subtract the sum of the numbers that are counted twice. The sum of the multiples of 3 is (3 + 6 + 9 + ... + 99 = \frac{33(100)}{2} = 1650), and the sum of the multiples of 5 is (5 + 10 + 15 + ... + 95 = \frac{20(100)}{2} = 1000). The sum of the numbers that are counted twice is (15 + 30 + 45 + ... + 90 = \frac{15(100)}{2} = 750). Therefore, the sum of all the elements of (S) is (1650 + 1000 - 750 = 1350).

The numbers that are divisible by 7 but not by 11 are the multiples of 7 that are not multiples of 11. The multiples of 7 less than 1000 are (7, 14, 21, ..., 994), and the multiples of 11 less than 1000 are (11, 22, 33, ..., 990). The numbers that are multiples of both 7 and 11 are (77, 154, 231, ..., 931). Therefore, the number of positive integers less than 1000 that are divisible by 7 but not by 11 is (\frac{994}{7} - \frac{990}{11} + \frac{931}{77} = 132).

The elements of (S) are the multiples of 2 and 3 less than 100. The multiples of 2 are (2, 4, 6, ..., 98), and the multiples of 3 are (3, 6, 9, ..., 99). However, some numbers are counted twice, such as 6, which is a multiple of both 2 and 3. To find the sum of all the elements of (S), we need to subtract the sum of the numbers that are counted twice. The sum of the multiples of 2 is (2 + 4 + 6 + ... + 98 = \frac{2(100)}{2} = 1000), and the sum of the multiples of 3 is (3 + 6 + 9 + ... + 99 = \frac{3(100)}{2} = 1500). The sum of the numbers that are counted twice is (6 + 12 + 18 + ... + 96 = \frac{6(100)}{2} = 300). Therefore, the sum of all the elements of (S) is (1000 + 1500 - 300 = 1800).

The elements of (S) are the multiples of 3 and 4 less than 100. The multiples of 3 are (3, 6, 9, ..., 99), and the multiples of 4 are (4, 8, 12, ..., 96). However, some numbers are counted twice, such as 12, which is a multiple of both 3 and 4. To find the sum of all the elements of (S), we need to subtract the sum of the numbers that are counted twice. The sum of the multiples of 3 is (3 + 6 + 9 + ... + 99 = \frac{3(100)}{2} = 1500), and the sum of the multiples of 4 is (4 + 8 + 12 + ... + 96 = \frac{4(100)}{2} = 2000). The sum of the numbers that are counted twice is (12 + 24 + 36 + ... + 96 = \frac{12(100)}{2} = 600). Therefore, the sum of all the elements of (S) is (1500 + 2000 - 600 = 1500).

To find the greatest integer less than 100 that is divisible by both 3 and 5, we can find the least common multiple (LCM) of 3 and 5, which is 15. The greatest integer less than 100 that is divisible by 15 is 90.

The GCD of 18 and 24 can be found using the Euclidean algorithm: 24 = 18 * 1 + 6; 18 = 6 * 3 + 0. Therefore, the GCD is 6.

To be divisible by 100, n! must contain at least two factors of 5. The smallest positive integer that satisfies this condition is 25, as 25! contains two factors of 5.

The multiples of 7 less than 100 are: 7, 14, 21, ..., 98. The sum of this arithmetic sequence can be calculated using the formula: sum = (n/2) * (a1 + an), where n is the number of terms, a1 is the first term, and an is the last term. Plugging in the values, we get: sum = (14/2) * (7 + 98) = 840.

To be divisible by 3 but not by 5, a number must be divisible by 3 but not divisible by 15. The multiples of 3 less than 1000 are: 3, 6, 9, ..., 999. The multiples of 15 less than 1000 are: 15, 30, 45, ..., 990. Subtracting the multiples of 15 from the multiples of 3, we get the numbers that are divisible by 3 but not by 5. There are 266 such numbers.

Using the Euclidean algorithm, we have: 36 = 24 * 1 + 12, 24 = 12 * 2 + 0. Therefore, the GCD of 24 and 36 is 12.