find out the squres of 25 consecutive odd numbers
Answers
Answer:
Step-by-step explanation:
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!
square-numbers
share cite improve this question follow
asked Jun 17 '14 at 17:53
user130520
11511 silver badge66 bronze badges
add a comment
4 Answers
1
We know that the sum of the odd numbers up to 2n−1 is n2. The sum of the odd numbers from 2m+1 through 2n−1 is then n2−m2=(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⋅15 (Why?) For example 225=45⋅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
share cite improve this answer follow
answered Jun 17 '14 at 18:26
Ross Millikan
348k2626 gold badges230230 silver badges415415 bronze badges
add a comment
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
share cite improve this answer follow
answered Jun 17 '14 at 18:01
DSinghvi
53533 silver badges1616 bronze badges
Shouldn't it be 2n−1, so for 25=52=1+3+5+7+9? – Ross Millikan Jun 17 '14 at 18:31
depends where you start the series But Yes I agree to your notion that someone might feel uncomfortable with it @RossMillikan – DSinghvi Aug 25 '14 at 17:04
add a comment
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
1×225=3×75=5×45=9×25=15×15=25×9=45×5=75×3=225×1
Five of those correspond to a sum of positive consecutive odd numbers
The sum of consecutive odd numbers from 1 to 2n−1 is n2
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=(33+12)2.