What is the product of all the values of n that make 546,324,16n divisible by 6
To solve this problem, we need to determine the values of n that make 546,324,16n divisible by 6.
We can start by writing the prime factorization of 6: 6 = 2 * 3.
Since 546,324,16n is divisible by 6, it must contain at least one factor of 2 and one factor of 3.
Considering the factor of 2 first, we can see that the last digit of the number must be even for it to be divisible by 2. So, n can be any even number from 0, 2, 4, 6, 8, 10, etc.
Now, considering the factor of 3, we know that a number is divisible by 3 if the sum of its digits is divisible by 3. To find values of n that satisfy this condition, we need to find the sum of digits in the number 546,324,16n.
Sum of digits = 5 + 4 + 6 + 3 + 2 + 4 + 1 + 6 + n = 31 + n
For the number to be divisible by 3, the sum of digits must be divisible by 3.
We need to find values of n for which (31 + n) is divisible by 3 or (n + 1) is divisible by 3.
Since n can be any even number from 0, 2, 4, 6, 8, 10, etc., we can see that the values of n that satisfy (n + 1) being divisible by 3 are 2, 8, 14, 20, and so on.
The product of these values can be calculated as:
2 * 8 * 14 * 20 = 4,480
Therefore, the product of all the values of n that make 546,324,16n divisible by 6 is 4,480.