four consecutive even entegers

Juan,

I believe that you have copied your problem incorrectly. There are no 4 consecutive EVEN integers that multiply out to 1680.

You can see this for yourself - lets pick the first 4 even integers: 2, 4, 6, 8. The product of these even integers is 384 - too small. The next potential set is 4, 6, 8, 10, and the product of these four is 1920, which is too big.

Therefore, you must have copied or typed the problem wrong.

Assuming that you didn't mean to say EVEN integers, there is a solution. If we assign the first number in the sequence to this variable 'n', then the four numbers would be:

n, n+1, n+2, n+3

The product of these is:

n(n+1)(n+2)(n+3) = 1680

Which we can FOIL. I've done this in three steps - first FOILing the first two terms 'n' and 'n+1' and FOILing 'n+2' and 'n+3', then FOILing the two results, and then combining terms and collecting them on the left side of the equation:

n(n+1)(n+2)(n+3) = 1680

(n^{2} + n)(n^{2} + 5n + 6) = 1680

(n^{4} + 5n^{3} + 6n^{2} + n^{3} + 5n^{2} + 6n) = 1680

n^{4} + 6n^{3} + 11n^{2} + 6n - 1680 = 0

Now we have to solve this equation for n. There are several ways to do this, none of which are easily described in this forum. I used a program called "Octave" to find the solutions; it is similar to Matlab. However, there are two REAL answers: n = -8 and n = 5.

So, ASSUMING that you are NOT looking for EVEN integers, but any integers, the answer is EITHER: (5, 6, 7, 8) OR (-8, -7, -6, -5). Multiplying either of these sets of numbers results in 1680. Notice that both sets of numbers are the same, except one is positive and one is negative. But since it is an even number of integers, the answers to the multiplication are both positive.

## Comments

Thank you for the comment on my mistake.

You don't need any program to do it.

1680^(1/4) ~ 6

So, the 4 consecutive numbers should be around 6.

5*6*7*8 = 1680

(-5)(-6)(-7)(-8) = 1680

Answer: {-8, -7, -6, -5} and {5, 6, 7, 8}