Suppose a1, a2, . . . , an is a list of n numbers with the following properties:
The sum of those n numbers is 500.
The sum of the smallest three of those numbers is 48.
The sum of the largest two of those numbers is 35.
(Note: There might be some repetitions among the numbers ak in that list.)
(a) What are the possible values of n ? Explain your reasoning.
(b) Can each of those values of n occur if we require all the numbers ak to be integers?
find two consecutive positive integer sum of the whose squares is 365
To find the possible values of n, let's analyze the given information step by step:
(a) The sum of all n numbers is 500. This means that the average value of each number is 500/n.
(b) The sum of the smallest three numbers is 48. Since the average value is 500/n, the sum of the smallest three numbers can be written as 3 * (500/n) = 1500/n.
(c) The sum of the largest two numbers is 35. Using the same logic, we can write this as 2 * (500/n) = 1000/n.
Based on the given information, we know that:
1500/n = 48 --> n = 1500/48 --> n ≈ 31.25
1000/n = 35 --> n = 1000/35 --> n ≈ 28.57
Since n represents the number of elements in the list, it must be a whole number. Therefore, n could be either 31 or 29.
Now let's move to part (b) of the question.
If we require all the numbers ak to be integers, we need to consider whether it's possible to find n numbers that satisfy all the given conditions.
For n = 31:
The sum of the smallest three numbers is 48, which means the average of those three numbers is 48/3 = 16. Since we need integers, one possibility is to have three numbers: 15, 16, and 17. Then, the remaining 28 numbers must have a sum of 500 - (15 + 16 + 17) = 452. There are multiple ways to distribute these 28 remaining numbers. For example, we can have 8 numbers equal to 16 and 20 numbers equal to 17.
For n = 29:
Using a similar approach, we can find that one possibility is having four numbers: 8, 9, 10, and 21 as the smallest four numbers. The remaining 25 numbers must have a sum of 500 - (8 + 9 + 10 + 21) = 452. Again, there are multiple ways to distribute these 25 remaining numbers.
In conclusion, both n = 31 and n = 29 are possible values if we require all the numbers ak to be integers.