Find all the ways to express 225 as a sum of consecutive odd integers

Question

Use your results to find the squares that can be added to 225 to produce another square.

I started off by taking the 9 divides 225 with quotient 25.

(25-8) + (25-6) + (25-4) + (25-2) + 25 + (25+2) + (25+4) + (25+6) + (25+8) = 225

simplifying:

17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 = 225.

Since integers can be negative, some ways will start with negative integers. For example 45 consecutive odd integers can add to 225 like this: (-39) + (-37) + (-35) + ... + 45 + 47 + 49 = 225.

So, to answer your question you have to count the number positive odd divisors of 225, and that will be the cardinality of the the set {1, 3, 5, 9, 15, 25, 45, 75, 225}, so the answer is 9.

I'm having trouble with finding the ways to express the sum of consecutive odd integers. I think I am off to a good start but really need a formal answer as well as a way to produce another square. Please help! Thanks!

Answer 1

Hint: Do you know a square can be written as sum of odd consecutive integer from one to 2n+1 where n= square root of number

Answer 2

Hint:

• the sum of a even number of odd numbers is even
• the sum of an odd number of odd numbers is odd
• the sum of an odd number of consecutive odd numbers is the number of numbers times the mean (or median) number
• $1 \times 225 = 3 \times 75 = 5 \times 45 = 9 \times 25 = 15 \times 15 = 25 \times 9 = 45 \times 5 = 75 \times 3 = 225 \times 1$
• Five of those correspond to a sum of positive consecutive odd numbers
• The sum of consecutive odd numbers from $1$ to $2n-1$ is $n^2$
• So there are five ways of filling the gap below the sum to $225$ with a non-negative sum of odd numbers

As an illustration $17 + 19 + 21 +23 +25 + 27+29+31+33=225$, to which you can add $1+3+5+7+9+11+13+15 = 64$ to give $289 = (\frac {33+1}{2})^2$.

Answer 3

We know that the sum of the odd numbers up to $2n-1$ is $n^2$. The sum of the odd numbers from $2m+1$ through $2n-1$ is then $n^2-m^2=(n+m)(n-m)$ We need this to be $225$, so you are looking for the number of solutions to $225=(n+m)(n-m)$ Each factorization of $225$ gives you one except $225=15\cdot 15$ (Why?) For example $225=45\cdot 5$ We can then write $n+m=45,n-m=5, n=25, m=20$ and find $225=41+43+45+47+49$. Note that there are $5$ terms in the expression and the middle one is $45$

Answer 4

$225=9\times25$ shown as the sum of $9$ consecutive odd integers.