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Title
Consecutive problem sum discussion

Problem Statement
The sum of five positive consecutive odd numbers is 3055. Find the least of these numbers.

Problem setup

We need to find the smallest of five consecutive odd numbers that total 3055 when added together.

Plans to Solve/Investigate the Problem

I personally enjoy working these types of problems now because when I was an algebra student, I could not make any sense of them.

There are two different strategies that I used to approach this problem.  The first strategy was numerically based.  There were more efficient ways to approach the "ballpark of probable solutions", but I initially guessed that the numbers must be in the 600's range, since 3,000 ÷  5 = 600.  The second strategy was an algebraically based solution, setting the first consecutive integer equal to a variable and increasing each of the following integers by 2, terminating at the fifth consecutive odd number.

Investigation/Exploration of the Problem

The first strategy was numerically based combined with "guess and check".  This method tends to be employed most by my middle school students that have not yet begun to exploring methods of creating personal algorithms or utilizing algebra.  I began with the smallest number as 691 followed by 693, 695, 697, and 699.  These totaled 3475.  Much too large.  I next approached a much smaller series of numbers beginning with 611 followed by 613, 615, 617, and 619.  These totaled 3077.  Closer but still too large. I continued with this strategy until I began to approach the desired solution from the lower end beginning with 601.  These consecutive integers totaled 3025.  I eventually approached the correct solution with 607 being the smallest number and the consecutive odd integers being 609, 611, 613 and 615 with a total of 3055.

I am most comfortable with the second strategy, an algebraic based solution.  I created the following variable based representation of five consecutive odd numbers totaling 3055:

x  =  first consecutive odd number

x  +  2  = second consecutive odd number  (if the x = 1, the next consecutive odd

odd number would be x  + 2 or

1 + 2 = 3)

x  +  4  = third consecutive odd number

x  +  6  = fourth consecutive odd number

x  +  8  =  fifth consecutive odd number

3055

Combining like terms,  (x)  +  (x + 2)  +  (x + 4)  +  (x + 6)  +  (x + 8)  =  3055

5x  +  20  =  3055

+     -20         -20       5x is isolated to the left of the equal sign by adding

negative 20 to both sides of the equation, remembering that

in order for equations to remain equal, what is done on one

side of the equal side must be done to the other side as well.

5x  =  3035                 The inverse operation of multiplication is division or as I tell

middle school children, "We undo multiplication with division."

5x  =  3035

5         5

5x  =  3035                We are left with x equal to 3035 ÷ 5 which equals 607.

5         5

The variable x was set to equal the smallest of the odd numbers, so the solution is 607

Extensions of the Problem

What elements would change about the problem if we were looking at 4 consecutive odd numbers as opposed to 5?

Author & Contact
Angela Gilliam
agilliam@rockdale.k12.ga.us

Link(s) to resources, references, lesson plans, and/or other materials
Consecutivenum