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
Link 2
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