altairahmadThe first thing the poster did was find the total count of ways that the Number 75 can be expressed as the Sum of N Consecutive Integers.
Generally, the number of ways of writing some constant M (here 75) as the SUM of N Consecutive Integers can be found by finding the Number or Odd Factors of M (here 75) EXCEPT the odd factor of 1
75 = (3) (5)^2
Total no. of odd factors = (1 + 1) * (2 + 1) = 2 * 3 = 6
6 - (1) = 5
There should be 5 ways to write 75 as the SUM of 2 or more consecutive positive integers.
Further, a good approach is the following:
M (here 75) can be expressed as the SUM of N consecutive integers if and only if either:
(1) the N consecutive integers is an ODD count, then: (M) / (N) must = Integer
Or
(2) if the N consecutive integers is an EVEN Count then: (M) / (N) + (1/2) must = Integer
The key constraint to avoid having zero or negative integers in your SUM is:
(M/N) >/= (N/2)—— otherwise we still start involving negative numbers along with zero
The cut off is when N = 13 consecutive positive integers
(75/13) < (13/2)
So we are looking for N to be any number from 2 through and including 12
Ignoring the “mathy” approach, I think the best way of approaching this problem is to form an organized “hit and run” approach.
I believe some people expressed this method above, but say you want to express 75 as the sum of 2 consecutive positive integers:
(75/2) = 37.5
37.5 would be the Median of the 2 consecutive integers. The 2 consecutive integers would then “sandwich” this value of 37.5 in the middle as the Median.
36 + 37 = 75
If you combine this approach with the knowledge that there are 5 ways to write 75 as the sum of consecutive positive integers (as found above), you can keep an organized trial approach until you find the 5th way.
Also, we know that in a set of consecutive positive integers, if there is an Even Count of terms, the Median must take the form of ——> XXX . 5
If there is an Odd count of terms in a set of consecutive positive integers, then the median must be one of the terms in the set and thus an Integer.
75/3 = 25 = Integer —— can do it
24 + 25 + 26 = 75
So far N = 2 and 3 works
75/4 = not of the form XX.5, so we won’t be able to make a sequence of positive consecutive integers around the result
75/5 = 15 —— Integer, so will work
13 + 14 + 15 + 16 + 17 = 75
N = 2, 3, and 5 work thus far - we have two more to find
75/6 = 12.5
10 + 11 + 12 + 13 + 14 + 15 = 75
N = 2, 3, 5, and 6 work - we have one more to find
75/7 ——> not an integer
75/8 ——> not of form XX . 5
75/9 —-> not an integer
75/10 = 7.5
3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 75
You can see at this point even if we went much further, we will start having to involve 0 and/or negative numbers in the set of consecutive positive integers.
The five ways to write 75 as the sum of consecutive positive integers are:
2—3—5—6—10
Sum of these = 2 + 3 + 5 + 6 + 10 = 26
Answer: 26
altairahmad wrote:
CaptainLevi wrote:
75 = \(3 X 5^2\)
no of odd factors = (2+1)(1+1)-1=5
Total five ways we can write 75 as sum of consecutive numbers
But question ask the value of sum of n
\((n)*(2a + (n-1))\) = 150 150 can be written as
2 X 75
3 X 50
5 X 30
6 X 25
10 X 15
The first number indicates the number of terms in that sequence example 3 x 50 implies 150 /3 = 25 implies 24+25+26
Therefore 2 + 3 + 5 + 6 + 10 = 26
Hi,
Why did you find the number of odd factors when we also have n=2, 6 and 10 i.e even number of consecutive terms.
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