Consecutive N

The sum of N consecutive positive integers is 1998. If N is odd, what is the greatest possible value for N?
Source: mathcontest.olemiss.edu 9/26/2011

SOLUTION
Before tackling this problem we should revisit two formulas related to the sum of consecutive positive integers. Let a_1,a_2,a_3,\cdots,a_N where N is odd be a collection of consecutive positive integers.
For example, 6,7, 8,9,10 is such a collection where N=5;a_1=6,a_2=7,\cdots,a_5=10

Sum formula
The formula that calculates the sum is given by
Sum=\frac{\left (a_N+a_1\right )\left (a_N-a_1+1\right )}{2}
The second factor in the above formula \left (a_N-a_1+1\right ) really is the number of terms N in the collection, so in this problem we have a simpler formula
Sum=\frac{\left (a_N+a_1\right )N}{2}
The sum in our example
6+7+8+\cdots+10=\frac{\left (10+6\right )5}{2}=40

Sum of first and last term
a_1=a_1
a_2=a_1+1
a_3=a_2+1=\left (a_1+1\right )+1=a_1+2
a_4=a_3+1=\left (a_1+2\right )+1=a_1+3
\cdots
a_N=a_{N-1}+1=\left (a_1+N-2\right )+1=a_1+\left (N-1\right )
Thus,
a_1+a_N=a_1+a_1+\left (N-1\right )
a_1+a_N=2a_1+\left (N-1\right )        (1)
In our example,
6+10=2\times 6+\left (5-1\right )=12+4=16

Guess and check method
Applying the sum formula yields
\frac{\left (a_N+a_1\right )}{2}\times N=1998
Decomposing 1998 into prime factors gives
1998=2\cdot 3^3\cdot 37

Case 1: N=3^3\cdot 37=999
\frac{\left (a_N+a_1\right )}{2}\times 999=1998
\frac{\left (a_N+a_1\right )}{2}=\frac{1998}{999}=2
a_N+a_1=4
Applying Eq. (1)
a_1+a_N=2a_1+\left (N-1\right )
4=2a_1+\left (999-1\right )
4=2a_1+998
4-998=2a_1
-994=2a_1
-497=a_1       
not a solution because negative integers are not allowed.

Case 2: N=3^2\cdot 37=333
\frac{\left (a_N+a_1\right )}{2}\times 333=1998
\frac{\left (a_N+a_1\right )}{2}=\frac{1998}{333}=6
a_N+a_1=12
Applying Eq. (1)
a_1+a_N=2a_1+\left (N-1\right )
12=2a_1+\left (333-1\right )
12=2a_1+332
12-332=2a_1
-320=2a_1
-160=a_1       
not a solution

Case 3: N=3\cdot 37=111
\frac{\left (a_N+a_1\right )}{2}\times 111=1998
\frac{\left (a_N+a_1\right )}{2}=\frac{1998}{111}=18
a_N+a_1=36
Applying Eq. (1)
a_1+a_N=2a_1+\left (N-1\right )
36=2a_1+\left (111-1\right )
36=2a_1+110
36-110=2a_1
-74=2a_1
-37=a_1       
not a solution

Case 4: N=37
\frac{\left (a_N+a_1\right )}{2}\times 37=1998
\frac{\left (a_N+a_1\right )}{2}=\frac{1998}{37}=54
a_N+a_1=108
Applying Eq. (1)
a_1+a_N=2a_1+\left (N-1\right )
108=2a_1+\left (37-1\right )
108=2a_1+36
108-36=2a_1
72=2a_1
36=a_1       
a solution!

Check
N=37;a_1=36;a_{37}=72
36+37+\cdots +72=\frac{\left (72+36\right )37}{2}
=\frac{\left (108\right )37}{2}
=\left (54\right )37
=1998

Answer: N=37.

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , . Bookmark the permalink.

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