Did you know that \(1225\) can be written as the sum of ten consecutive integers?
That is, \[1225=118+119+120+121+122+123+124+125+126+127\]
The notation below illustrates a mathematical short form used for writing the above sum. This notation is called Sigma Notation.
\[\sum_{i=118}^{127}i = 1225\]
How many ways can the number \(1225\) be expressed as the sum of an odd number of consecutive positive integers?
We will use the following idea to solve this problem: If there exists an odd number, \(k\), of consecutive integers that sum to \(1225\), then \(k\) is a divisor of \(1225.\)
Furthermore, if \(kn=1225\), then \(n\) is the mean (average) of the \(k\) integers, and will appear in the middle of the sequence of numbers summing to \(1225.\)
Why is this true? Let’s first consider \(k=5.\)
Five consecutive integers can be expressed as \(n-2\), \(n-1\), \(n\), \(n+1\), and \(n+2\), where \(n\) is an integer.
Their sum is \((n-2)+(n-1) + n +
(n+1)+(n+2)=5n.\)
Therefore, \(5n=1225\) and \(n=245\). Thus, the middle term in the sum
is \(245\), and the series is \(243+244+245+246+247 = 1225.\)
In general, if there are \(k\) consecutive integers, where \(k\) is odd, and \(n\) is the middle number in the sum, then there are \(\frac{k-1}{2}\) integers less than \(n\) in the sum and \(\frac{k-1}{2}\) integers greater than \(n\) in the sum. Furthermore, the first integer in the sum is \(n-\frac{k-1}{2}\), the last integer in the sum is \(n+\frac{k-1}{2}\), and we can write the sum of these integers in this way: \[\left(n-\frac{k-1}{2}\right)+ \cdots + (n-3) + (n-2) + (n-1) + n + \space \]\[ (n+1) + (n+2) + (n+3) + \cdots + \left(n+\frac{k-1}{2}\right)\]
This simplifies to \(kn\). Thus, if this sum is equal to \(1225\), then \(kn=1225\) and so \(k\) is an odd divisor of \(1225.\)
Since \(1225=5^27^2\), the positive divisors of \(1225\) are \(1\), \(5\), \(7\), \(25\), \(35\), \(49\), \(175\), \(245\) and \(1225\), which are all odd.
For each odd divisor, \(k\), of \(1225\), we determine \(n= \frac{1225}{k}\), which will be the middle term in the sum. The \(k\) integers that sum to \(1225\) will then be \(\left(n-\frac{k-1}{2}\right) + \cdots + n + \cdots+\left(n+\frac{k-1}{2}\right)\). This is summarized in the table below.
| Number of Integers (\(k\)) | Middle Integer (\(n\)) | Sum of Integers |
|---|---|---|
| \(1\) | \(1225\) | \(1225\) |
| \(5\) | \(245\) | \(243+244+245+246+247\) |
| \(7\) | \(175\) | \(172+173+174+175+176+177+178\) |
| \(25\) | \(49\) | \(37+38+\cdots+49+\cdots+60+61\) |
| \(35\) | \(35\) | \(18+19+\cdots+35+\cdots+51+52\) |
| \(49\) | \(25\) | \(1+2+\cdots+25+\cdots+48+49\) |
| \(175\) | \(7\) | \((-80)+(-70)+\cdots+7+\cdots+93+94\) |
| \(245\) | \(5\) | \((-117)+(-116)+\cdots+5+\cdots+126+127\) |
| \(1225\) | \(1\) | \((-611)+(-610)+\cdots+1+\cdots+612+613\) |
Note that all integers in the sum are positive for \(k = 1, 5, 7, 25, 35, 49.\) For \(k=175, 245, 1225\), there are negative integers in the sum.
Thus, there are six ways to express \(1225\) as the sum of an odd number of consecutive positive integers.
Extension: Determine the number of ways the number \(1225\) can be expressed as the sum of an even number of consecutive positive integers.