How many numbers
from 1 to 1000
inclusive can be
expressed as the sum
of k >= 2 consecutive
positive integers for
some value of k ? Sorry to post question from Brilliant, I got 997 integers, but it's wrong. If you think you can answer ...
Hmmm.
For k=2, we want all n<=1000 such that n=m+m+1 = 2m+1
That is, all odd numbers 3<=n<=999
There are 996/2+1 = 499 of those
For k=3, we want n=3m+3
so, that's all multiples of 3 with 6<=n<=999
There are 993/3+1 = 332 of those
For k=4, we want 4m+6
so, that's all multiples of 4 with 10<=n<=998
There are 988/4+1 = 248 of those
There's a pattern here, so you should be able to come up with some
∑
k=2
expression, or just brute-force your way through.
Maybe not a lot of help, but it does show that there are more than 997 of them.
The ans Steve gave is absolutely wrong...the ans is less than 823 and greater than 665...steve u have counted the same numbers more than one time...for example u have counted 15 twice....
When I calculated out I got 990 as the correct answer.
To find the number of integers from 1 to 1000 inclusive that can be expressed as the sum of k >= 2 consecutive positive integers, we can follow these steps:
1. First, we need to understand the pattern of how integers can be expressed as the sum of consecutive positive integers.
2. Let's consider an example. To find the consecutive positive integers that sum up to 15, we can start with 1 and 2 as the first two consecutive integers. If we keep adding the next consecutive positive integers, we get the following sums:
- 1 + 2 = 3
- 1 + 2 + 3 = 6
- 1 + 2 + 3 + 4 = 10
- 1 + 2 + 3 + 4 + 5 = 15
3. Notice that the number of consecutive positive integers we need to sum up to a specific number is equal to the square root of twice that number, rounded down to the nearest integer.
- For example, in the case of 15, the square root of (2 * 15) is approximately 5.48, and rounded down to the nearest integer, we get 5.
- So, if we have a number n, we can find the maximum value of k (the number of consecutive positive integers) by calculating sqrt(2 * n) and rounding it down.
4. Now, we can iterate through the numbers from 1 to 1000 and check how many of them can be expressed as the sum of k consecutive positive integers, where k ranges from 2 to the maximum value calculated in step 3.
5. For each number, we start with k=2 and iterate until k reaches the maximum value. We check if the number can be expressed as the sum of k consecutive positive integers.
- We can use a nested loop. The outer loop iterates from 2 to the maximum value we calculated.
- The inner loop starts with the first number in the consecutive series and iterates until the sum of the consecutive numbers is greater than or equal to the target number.
- If the sum of the consecutive numbers is equal to the target number, we increment a counter variable by 1, indicating that we found one integer that satisfies the condition.
6. Finally, we output the value of the counter variable, which represents the total number of integers from 1 to 1000 that can be expressed as the sum of k consecutive positive integers.
Following these steps, let's calculate the number of integers that can be expressed as the sum of k consecutive positive integers.