3 consecutive numbers multiply to 4080. What are the numbers?
Let the middle number be x.
(x-1)x(x+1) = x^3 - x = 4080
x = 16 works
I did not solve the cubic equation, but knew the answer had to be an integer slightly larger than the cube root of 4080.
The numbers are 15, 16 and 17.
To find the three consecutive numbers that multiply to 4080, we can use trial and error or set up an equation. Let's use the trial and error method:
Let's assume the three consecutive numbers are x, x+1, and x+2.
Now, we can set up an equation:
x * (x+1) * (x+2) = 4080
Expanding this equation:
x^3 + 3x^2 + 2x = 4080
Rearranging the terms:
x^3 + 3x^2 + 2x - 4080 = 0
By solving this equation, we can find the value of x. However, finding the exact value of x using this equation might be quite complex. Instead, let's try some values of x through trial and error.
Let's start by assuming x as 15:
15 * 16 * 17 = 4080
This satisfies our condition, and the three consecutive numbers are 15, 16, and 17.
To find the three consecutive numbers that multiply to 4080, we can solve this problem algebraically. Let's assume the first number is a, then the second number would be a+1, and the third number would be a+2.
Then we can set up the equation:
a * (a + 1) * (a + 2) = 4080
Expanding this equation, we get:
(a^2 + a)(a + 2) = 4080
a^3 + 3a^2 + 2a = 4080
a^3 + 3a^2 + 2a - 4080 = 0
Now, we can solve this equation either by factoring or by using numerical methods such as trial and error or a calculator.
Let's solve this equation using numerical methods:
By trying different values, we find that a = 14 satisfies the equation. Substituting a = 14 into the equation, we get:
14^3 + 3(14)^2 + 2(14) - 4080 = 0
2744 + 588 + 28 - 4080 = 0
Now, we can see that the equation balances out. Therefore, the three consecutive numbers that multiply to 4080 are 14, 15, and 16.
So, the numbers are 14, 15, and 16.