The product of the first ten prime numbers must be divisible by:
F. 16
G. 18
H. 20
J. 22
K. 24
Please provide detailed explanation with answer (if possible)
Thanks in advance.
well the first 10 prime numbers are 2,3,5,7,11,13,17,19,23,29.
The product of two of the primes, 2 and 11 is 22.
Therefore the product of the first ten primes which contains 2 and 11 will be 22.
So J (22) would be your answer.
To find out which one of the options, F, G, H, J, or K, the product of the first ten prime numbers is divisible by, we need to determine the prime factorization of each of the options and compare it with the prime factorization of the product of the first ten prime numbers.
The first ten prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
To find the prime factorizations of the options, we need to break them down into their prime factors:
F. 16 = 2^4
G. 18 = 2 * 3^2
H. 20 = 2^2 * 5
J. 22 = 2 * 11
K. 24 = 2^3 * 3
Now, let's calculate the prime factorization of the product of the first ten prime numbers:
Product = 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29
When we analyze the factors of the product of the first ten prime numbers, we can notice that it contains 2 raised to the power of 8, 3 raised to the power of 2, 5 raised to the power of 1, 7 raised to the power of 1, 11 raised to the power of 1, 13 raised to the power of 1, 17 raised to the power of 1, 19 raised to the power of 1, 23 raised to the power of 1, and 29 raised to the power of 1.
So, the product of the first ten prime numbers is divisible by:
F. 16 because 16 = 2^4, and the product contains 2 raised to the power of 8.
G. 18 because 18 = 2 * 3^2, and the product contains 2 raised to the power of 8 and 3 raised to the power of 2.
H. 20 because 20 = 2^2 * 5, and the product contains 2 raised to the power of 8 and 5 raised to the power of 1.
J. 22 because 22 = 2 * 11, and the product contains 2 raised to the power of 8 and 11 raised to the power of 1.
K. 24 because 24 = 2^3 * 3, and the product contains 2 raised to the power of 8 and 3 raised to the power of 2.
Therefore, the correct answer is that the product of the first ten prime numbers is divisible by options F, G, H, J, and K.
To solve this problem, we need to find the product of the first ten prime numbers and determine if it's divisible by each given number. Let's break it down step by step:
Step 1: Find the first ten prime numbers.
Prime numbers are numbers that are only divisible by 1 and themselves. The first ten prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
Step 2: Find the product of the first ten prime numbers.
To find the product, we multiply all these prime numbers together:
2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29 = 6469693230
Step 3: Test divisibility by each given number.
F. 16: To determine if 6469693230 is divisible by 16, we need to check if it's divisible by 2 four times (since 16 = 2^4). If the last four digits of the number are divisible by 2, then the whole number is divisible by 16. Checking the last four digits of 6469693230 (i.e., 3230) shows that it is divisible by 2, so it is divisible by 16.
G. 18: To determine if 6469693230 is divisible by 18, we need to check if it's divisible by both 2 and 9. We've already established that it's divisible by 2. To check divisibility by 9, we need to add up all the digits of the number and see if the sum is divisible by 9. The sum of the digits in 6469693230 is 6+4+6+9+6+9+3+2+3+0 = 48, which is divisible by 9. Therefore, 6469693230 is divisible by 18.
H. 20: To determine if 6469693230 is divisible by 20, we need to check if it's divisible by both 2 and 10. We've already established that it's divisible by 2. Checking the last digit of the number (i.e., 0) shows that it is divisible by 10, so it is divisible by 20.
J. 22: To determine if 6469693230 is divisible by 22, we need to find the remainder when dividing by 22. If the remainder is 0, then the number is divisible by 22. Performing the division, we find that 6469693230 divided by 22 gives us a remainder of 16, so it is not divisible by 22.
K. 24: To determine if 6469693230 is divisible by 24, we need to check if it's divisible by both 2 and 3. We've already established that it's divisible by 2. To check divisibility by 3, we need to check if the sum of the digits is divisible by 3. As we found earlier, the sum of the digits is 48, which is divisible by 3. Therefore, 6469693230 is divisible by 24.
Answer: The product of the first ten prime numbers (6469693230) is divisible by the following numbers: F. 16, G. 18, H. 20, and K. 24.