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## consecutive numbers

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### consecutive numbers

by klaud » Thu Mar 15, 2012 9:39 am
EDITED

If x,y and z are three consecutive odd numbers and x^2+y^2+z^2=683, what is their sum?
Last edited by klaud on Thu Mar 15, 2012 11:44 am, edited 1 time in total.

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by nitin_stmicro » Thu Mar 15, 2012 10:43 am
3 numbers 12,13,14 total of their squares is 509 and for 13,14,15 it is 590.

there are no three consecutive numbers with squares adding upto 515.

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by klaud » Thu Mar 15, 2012 11:15 am
ok but how to find out that the sum of 12,13,14 squared is 590?

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by pemdas » Thu Mar 15, 2012 12:01 pm
klaud you seem to be lazier than myself
here's the trick 12^2=144 and 13^2= 144+(12+13)
14^2=13^2+(13+14) or 144+(12+13) +(13+14)
this formula works for every consecutive number squared, e.g. 15^2=14^2 +(14+15)
klaud wrote:ok but how to find out that the sum of 12,13,14 squared is 590?
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by GMATGuruNY » Fri Mar 16, 2012 9:37 am
klaud wrote:EDITED

If x,y and z are three consecutive odd numbers and x^2+y^2+z^2=683, what is their sum?
Since 20Â²+20Â²+20Â² = 1200, and the required sum here is 683, we know that x, y and z are each less than 20.

The units digit of 683 is 3.
Focus on the units digits of the 3 consecutive odd integers.
Options for the units digits are 1, 3, 5, 7, 9, 1, 3...
The squares of these digits are 1, 9, 25, 49, 81, 1, 9...
Only 9+5+9 = 23 will yield a units digit of 3.
Thus, the 3 integers are 13, 15 and 17.

13+15+17 = 45.

If this question appeared on the GMAT, we could plug in the answers.
When numbers are evenly spaced, median = average.

Median = 45/3 = 15, implying that the 3 integers are 13, 15 and 17.
13Â² + 15Â² + 17Â² = 169 + 225 + 289 = 683.
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by garryrother » Sat Mar 17, 2012 1:59 pm
Hi Folks,

Confused a bit !
What about -13, -15 and -17?

Am i missing anything overhere?

Thanks!

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by jzw » Sun Mar 18, 2012 10:35 am
garryrother wrote:Hi Folks,

Confused a bit !
What about -13, -15 and -17?

Am i missing anything overhere?

Thanks!
You would hypothetically need to consider negatives in a consecutive integer question if you saw one in the answer choices.

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by [email protected] » Sun Mar 18, 2012 9:00 pm
Without picking numbers it can be solved algebraically as well.
Let x be the smallest of the odd numbers <y, so y=x+2 and z>y so z=x+4 {SInce x,y,z are consecutive odd numbers}.
Then,
x^2+(x+2)^2+(x+4)^2=683
or, x^2+x^2+4x+4+x^2+8x+16=683
or, 3x^2+12x=663
or x^2+4x-221=0
or (x+17)(x-13)=0

Hence x is either 13 or -17
So y is either 15 or -15 and z is either 17 or -13.

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by killer1387 » Sun Mar 18, 2012 9:38 pm
klaud wrote:EDITED

If x,y and z are three consecutive odd numbers and x^2+y^2+z^2=683, what is their sum?
let me share my solution:

let consecutive odd be n,n+2, n+4
hence
n^2+ (n+2)^2+ (n+4)^2=683
n^2+4n-221=0
hence
n(n+4)=221=17*13
hence n=13, n+2=15, n+4=17

hence 13+15+17= 45

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by icemanKK » Sat Mar 24, 2012 6:38 am
Just a small suggestion to some of the solutions stated above.

We can also use n-2,n,n+2 ... this will help in cancelling out terms and will be marginally quicker.

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