In a finite list of positive integers, each number, except for the last number, is a proper divisor of the next number in the list. If the first number in the list is 1 and the last number is 650, what is the greatest possible sum of the numbers in the list?

We are given a finite list of positive integers where each number (except for the last number) is a proper divisor of the next number. The first number is 1, and the last number is 650. Our goal is to find the greatest possible sum of the numbers in this list.

### Key Conditions
- Each number, except the last, is a **proper divisor** of the next number.
- The first number is 1, and the last number is 650.
- We want to maximize the sum of the numbers in the list.

### Step-by-step approach:

1. **Divisors of 650**:
   We begin by identifying all the divisors of 650. The numbers in the list must be divisors of 650, as each number is a divisor of the next, and the last number is 650.

   First, we find the prime factorization of 650:
   \[
   650 = 2 \times 5^2 \times 13
   \]
   The divisors of 650 are all products of these prime factors:
   \[
   \text{Divisors of } 650: 1, 2, 5, 10, 13, 25, 26, 50, 65, 130, 325, 650
   \]

2. **Building the Sequence**:
   To maximize the sum of the numbers in the list, we should include as many divisors as possible, while ensuring that each number is a proper divisor of the next one.

   Since 1 must be the first number and 650 the last, we aim to construct the longest chain of divisors between 1 and 650.

3. **Choosing the Sequence**:
   Starting with 1, we can choose the next number as 2 (the smallest divisor greater than 1). We continue this process by choosing divisors that are larger but maintain the property of being proper divisors of the next.

   One possible chain with the greatest number of terms is:
   \[
   1 \to 2 \to 10 \to 50 \to 650
   \]

   - \(1\) is a divisor of \(2\),
   - \(2\) is a divisor of \(10\),
   - \(10\) is a divisor of \(50\),
   - \(50\) is a divisor of \(650\).

4. **Sum of the Sequence**:
   The sum of the numbers in this sequence is:
   \[
   1 + 2 + 10 + 50 + 650 = 713
   \]

### Conclusion:
The greatest possible sum of the numbers in the list is \( \boxed{713} \).

### Do you have any questions or would you like further details? Here are some related questions for exploration:

1. What if the last number was different, such as 1000? How would the solution change?
2. Can you prove that no larger sum is possible with a different sequence of divisors?
3. What is the effect of excluding certain divisors on the maximum possible sum?
4. How does the prime factorization of 650 play a role in this problem?
5. How would the problem change if the first number was not 1?

#### Tip: Always check for proper divisors carefully to ensure that each number satisfies the conditions of the problem!

Discover how to solve a number theory problem where each number in a list is a proper divisor of the next, starting with 1 and ending with 650. This solution involves divisors, prime factorization, and constructing a sequence for the greatest possible sum. Suitable for students in Grades 10-12.

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