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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!

user130520
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4 Answers4

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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 $45$

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

Steve Kass
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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

DSinghvi
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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 $289 = (\frac {33+1}{2})^2$.

Henry
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  • are these the ways to find the answer? i understand the hints but I'm not sure how to answer the question with them. – user130520 Jun 17 '14 at 18:10
  • Note the nine sums of consecutive odd numbers adding up to $225$. Throw away the four with negative numbers. The remaining five point to solutions by filling the gap to zero below the sum. So $225=15\times 15$ points to $0^2+225=15^2$ while $225 = 9 \times 25$ points to $8^2+225=17^2$ and $225=1\times 225$ points to $112^2+225=113^2$. – Henry Jun 18 '14 at 00:15