Question 1154151: Find three consecutive positive odd integers such that the sum of their squares is 371.
Found 5 solutions by ramkikk66, mananth, Alan3354, ikleyn, greenestamps:
Answer by ramkikk66(644)   (Show Source):
Answer by mananth(16949)  (Show Source):
You can put this solution on YOUR website! Find three consecutive positive odd integers such that the sum of their squares is 371.
let n be the first positive odd integer
second will be n+2
Third will be n+4
n^2 +(n+2)^2 +(n+4)^2=371
n^2 +n^2 +4n +4 + n^2 +8n +16=371
3n^2+12n + 20=371
3n^2+12n-351=0
divide by 3
n^2+4n-117=0
(n+13)(n-9)=0
n=-13 Or n=9
but n is positive
Therefore n=9
The numbers are 9,11,13
CHECK
9^2+11^2+13^2=371
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Find three consecutive positive odd integers such that the sum of their squares is 371.
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371/3 = 123.666...
The nearest square is 121 = 11^2
---> 11 is the middle integer
Answer by ikleyn(53618)  (Show Source):
You can put this solution on YOUR website! .
Find three consecutive positive odd integers such that the sum of their squares is 371.
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I will show another way to calculate.
It requires less calculations and is much more educative than shown by the other person.
Let n be the central number, so the numbers are (n-2), n and (n+2).
Our equation is
(n-2)^2 + n^2 + (n+2)^2 = 371.
Simplify and find 'n'
(n^2 - 4n + 4) + n^2 + (n^2 + 4n + 4) = 371
3n^2 + 8 = 371
3n^2 = 371 - 8 = 363
n^2 = 363/3 = 121
n =  = +/- 11.
We are looking for positive integers, so n = 11, and the numbers are 9, 11, 13.
Solved.
It is how this problem is EXPECTED to be solved.
Answer by greenestamps(13295)  (Show Source):
You can put this solution on YOUR website!
For a formal algebraic solution, look at the response from tutor @ikleyn.
Note that, very often, solving a problem involving an odd number of consecutive integers (or involving an odd number of numbers from ANY arithmetic sequence) is easiest if you let your variable represent the MIDDLE number in the sequence instead of the first. In this problem, observe in her solution that many terms of the resulting polynomials cancel, leaving a much simpler polynomial to work with.
And if a formal algebraic solution is not required -- as in a competitive math exam where the speed of solving the problem is important -- a quick solution can be found using logical reasoning.
The sum of the squares of the three consecutive positive odd integers is 371, so the square of the middle one should be approximately 371/3, or about 123. Since 11 squared is 121, it is almost certain that the middle of the integers is 11. Quick mental arithmetic then confirms that 9^2 + 11^2 + 13^2 = 371.
ANSWER: 9, 11, and 13
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