Question

the product of two consecutive positive even numbers is 1520. What are the numbers? when I try "n" and "n+2" i get an odd number

Answer 1

you can express those consecutive even numbers like "n" and "n+2"

and their product is (n)(n+2) = 1520

Answer 2

guess and check for this one
there is no other way

Answer 3

\(40\times 42=1680\) nope, too big. try again with smaller numbers

Answer 4

That's not quite the right equation...

Answer 5

@Albert0898 it is the PRODUCT not the sum

Answer 6

you cannot use an equation to solve this, only to rewrite the question, don't be fooled
you just have to grind it until you find it

Answer 7

38 and 40

Answer 8

satellite was close ;)

Answer 9

hahah, sorry i used your guess to approximate.

Answer 10

yeah a calculator helps
i started too high, went down a couple, got it in try two

Answer 11

n(n+2) = 1520
n^2 + 2n - 1520 = 0
(n-38)(n+40) = 0
Since the numbers are positive, the 2nd solution doesn't work.

n must be 38 and n + 2 is 40.

Answer 12

(I couldn't factor that as easily as I made it sound... so a calculator helped me too :) )

Answer 13

You definitely do not have to grind it out and guess. But the equation is not shown correctly above. It is (2N)(2N + 2) = 1520. Becomes a simple quadratic. Solve for N. Answer will be the 2N and 2N + 2.

Answer 14

so my question is, how did you know how to factor ? you had to come up with two numbers whose product is 1520 and that are two apart
in other words just rewrote the question, still had to guess and check

Answer 15

x = even number
x + 2 = consecutive even number

x + 2 * x = 1520

x + 2 = 1520/x

x = 1518/x

x * x = 1518

Find the square root of 1518.

Approximate and Closest Even Number: 38.

x = 38

x + 2 = 40

Answer 16

i guess if you wanted to be ridiculous you could write
\[x^2+2x=1520\] and complete the square
\[(x+1)^2=1521\]
\[x+1=\sqrt{1521}=39\]\[x=38\] but that is a silly thing to do

Answer 17

@satellite73 Yes, I agree with you... that's what I meant... you either guess and check to factor, or just guess and check vs. 1520. My only point was that it is possible to write an equation to solve something like this.

Answer 18

@JakeV8 yes you are right, but i maintain that changing a problem like this in to an equation and then solving it by solving the original problem in words is a dog chasing its tail

Answer 19

You guys are missing it all together. Set the variables as 2N and 2N + 2. If you set N to a positive integer, 2N and 2N + 2 are automatically even. Becomes a simple quadratic. See my previous post.

Answer 20

thanks all I think I found my problem

Answer 21

Ok, I'm here. Give me a second to refresh myself on this.

Answer 22

Ok, I'm up to speed. I still stand firmly on my answer methodology. The way to GUARANTEE that you have two even numbers is to use (2N)(2N + 2) = 1520 because 2N and 2N + 2 will be FORCED to be even if N is a positive integer. Absolutely. I strongly suggest you use this methodology. If I understand, your current question has something to do with 3 positive even numbers now? Is that correct?

Answer 23

And please, whatever you do, don't follow satellite73 on this problem. He is WAY off base suggesting that you guess. That's beyond ludicrous.

Answer 24

@tcarroll010, that is the equation I was using however, i am still unable to find two positive even numbers. No I do not have to find three evennumbers, I was seeing if the way I was working it would give even numbers for three. hope that makes sence, sorry my computer is slow. . I only need two positive even numbers