Answer :
Let x = number 1
Let (x+1) = number 2
Let (x+2) = number 3
[tex]x(x+1)(x+2)=120\\ (x^{2} +x)(x+2)=120\\ x^{3}+2 x^{2} + x^{2} +2x-120=0\\ x^{3} +3 x^{2} +2x-120=0\\ [/tex]
4 is a solution
[tex]x^{3} +3 x^{2} +2x-120=0\\ 4^{3} +3 (4)^{2} +2(4)-120=0\\ 64+48+8-120=0[/tex]
The first number is 4, the next two are 5 and 6
4x5x6=120
4+5+6=15
Let (x+1) = number 2
Let (x+2) = number 3
[tex]x(x+1)(x+2)=120\\ (x^{2} +x)(x+2)=120\\ x^{3}+2 x^{2} + x^{2} +2x-120=0\\ x^{3} +3 x^{2} +2x-120=0\\ [/tex]
4 is a solution
[tex]x^{3} +3 x^{2} +2x-120=0\\ 4^{3} +3 (4)^{2} +2(4)-120=0\\ 64+48+8-120=0[/tex]
The first number is 4, the next two are 5 and 6
4x5x6=120
4+5+6=15
let's call the first number x. the two numbers after it are x+1 and x+2. their product is 120.
x(x+1)(x+2) = 120
x(x^2 + 3x + 2) = 120
x^3 + 3x^2 + 2x = 120
x^3 + 3x^2 + 2x -120 = 0
whoa... that's a pretty nasty cubic... I'm not even sure how to go about that.
Never mind. Ignore all that stuff. Here's what we'll do.
Factor 120 to figure out what numbers could multiply to it:
120 = 5*3*2*2*2
We can break up that prime factorization into 2*2 * 5 * 2*3, which is the same thing as 4 * 5 * 6.
And there you go! Those are your three consecutive natural numbers that multiply to 120. Now find their sum:
4 + 5 + 6 = 15
x(x+1)(x+2) = 120
x(x^2 + 3x + 2) = 120
x^3 + 3x^2 + 2x = 120
x^3 + 3x^2 + 2x -120 = 0
whoa... that's a pretty nasty cubic... I'm not even sure how to go about that.
Never mind. Ignore all that stuff. Here's what we'll do.
Factor 120 to figure out what numbers could multiply to it:
120 = 5*3*2*2*2
We can break up that prime factorization into 2*2 * 5 * 2*3, which is the same thing as 4 * 5 * 6.
And there you go! Those are your three consecutive natural numbers that multiply to 120. Now find their sum:
4 + 5 + 6 = 15
