Can you find three consecutive even numbers that total 85008 when multiplied together ?

Answer 1

The way to solve this would be to let the 2nd consecutive number be x. So the three even consecutive numbers are x-2, x and x+2.

(x-2)(x)(x+2)=85008

This equation can be difficult to solve, but a short method would be to just cube root 85008 to get an approximate middle number.

Cube root (85008) = 44

Try multiplying 42x44x46 = 85008 (voila!)

The numbers are 42, 44 and 46.

Answer 2

I'd start with the cube root of 65008 and try the numbers around there. It turns out that cube root of 65008 is approximately 43.9696761, so try 44 for the middle number.

Lo and behold:

The answer is: 42 * 44 * 46 = 85,008

(Note: You could go the long way and end up with a third order polynomial and solve for n... but the shortcut method is much
easier.)

(n-2)n(n+2) = 85008
n(n² - 4) = 85008
n^3 - 4n = 85008
n^3 = 85008 - 4n
n = cube root (85,008 - 4n)

That's why you can take the cube root of 85,008 and instead use the closest integer value.

Answer 3

42, 44, 46

Answer 4

when you factor x^3+6x^2+8x-85008 = 0 you get

(x-42)(x^2+48x+2024)=0 which i got by using this site: http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=factor&s3=basic

I like puzzlings way of taking the cube root but this might be for school and they might want you to have the polynomial and factor it.

Answer 5

letthe numbers be n,n+2 and n+4
n(n+2)(n+4)=85008
(n^2+2n)(n+4)=85008
n^3+6n^2+8n=85008
the nos are 42,44 and 46

Answer 6

42,44,46