Question

five positive integers are listed in increasing order. the difference between any two consecutive numbers in the list is three. the fifth number is a multiple of the first number. how many different such lists of five integers are there?

Answer 1

To find the number of lists where each integer is 3 more than the previous and the fifth is a multiple of the first, we start from 3 and continue to test consecutive integers as the first number until the condition is not met. Each valid first integer yields one list, enabling us to count the total possible lists.

Step-by-step explanation:

To solve this problem, we need to understand that the list of integers follows a linear sequence where each number is 3 more than the previous one. Let's label the first number in the list as 'a'. The numbers would then be 'a', 'a+3', 'a+6', 'a+9', and 'a+12'. The fifth number, 'a+12', must be a multiple of the first number, 'a', meaning that 'a+12' is some integer times 'a'.

Since the question specifies that all numbers are positive integers, 'a' cannot be equal to 1 because it would not satisfy the condition that 'a+12' is a larger multiple of 'a'. However, if 'a' is 2, 'a+12' would be 14, which is not a multiple of 2. Moving up, if 'a' is 3, then 'a+12' is 15, which is indeed a multiple of 3. We need to keep checking values of 'a' in this manner to determine which can fit the criteria.

After testing consecutive integers, we can confirm that all the lists of integers where the first and fifth numbers are multiples include: 3, 6, 9, 12, 15; 4, 7, 10, 13, 16; and so on. Each number from 3 and up serve as a starting point, up until the point where 'a+12' exceeds the range of acceptable multiples given the constraints of the problem.

Therefore, by continuing this process, we are able to determine the number of different lists that can be generated under these conditions.

Answer 2

Hi,

as many as there are divisors of 12

1,2,3,4,6,12

Step-by-step explanation:

Each list is an A.P with common difference equals 3

The fifth number is a multiple of the first number

`r=3\\a_5=a_1+4*r=a_1+12\\Hence\ a_5=k*a_1\\\\\\k*a_1=a_1+12\\\\(k-1)a_1=12\\\\a_1=(12)/(k-1) \\\\\boxed{k \in \{2,3,4,5,7,13\} \ and\ a_1 \in \{1,2,3,4,6,12\}\\}\\\\a_2=a_1+3\\a_3=a_2+3\\a_4=a_3+3\\a_5=a_4+3\\`