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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consecutive numbers
Source: Beat The GMAT — Problem Solving |
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)
care to learn more short-cuts send me a private message (pm)
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)
care to learn more short-cuts send me a private message (pm)
klaud wrote:ok but how to find out that the sum of 12,13,14 squared is 590?
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Since 20²+20²+20² = 1200, and the required sum here is 683, we know that x, y and z are each less than 20.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?
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.
Answer choice C: 45
Median = 45/3 = 15, implying that the 3 integers are 13, 15 and 17.
13² + 15² + 17² = 169 + 225 + 289 = 683.
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You would hypothetically need to consider negatives in a consecutive integer question if you saw one in the answer choices.garryrother wrote:Hi Folks,
Confused a bit !
What about -13, -15 and -17?
Am i missing anything overhere?
Thanks!
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.
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.
let me share my solution: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 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
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.
We can also use n-2,n,n+2 ... this will help in cancelling out terms and will be marginally quicker.
