Question

Five consecutive even integers have a sum of -20. Quantity A = The greatest of the 5 integers. Quantity B= 0. Is it:
A) Quantity A is greater
B) Quantity B is greater
C) They are equal
D) Not enough info to determine.

Answer 1

Ok.

Answer 2

x (greatest of the five)
x+(x-2)+(x-4)+(x-6)+(x-8)=-20
Solve for x

Answer 3

Or, you could do a faster way

Answer 4

@RaphaelFilgueiras , I do believe that is unecessarily nasty.

Answer 5

how you can you say that x cannot be even?

Answer 6

No. x is even.

Answer 7

x is even because it is the largest of the sequence, and is defined to be even.

Answer 8

any consecutive even number has a difference of 2 from the previous number.

Answer 9

i read odd,but you don't have to put 2K?

Answer 10

Five consecutive even integers have a sum of -20
I assumed that x was the largest number.
Then, x must be even, because I assumed it was part of the sequ3ence.

Answer 11

i dont know is think yoy have to write this way
(2k)+(2k+2)+(2K+4)+(2k+6)+(2k+8)=−20

Answer 12

ok i got it!

Answer 13

but if we have 4...

Answer 14

@RaphaelFilgueiras, usually, in a proof involving even integers, you have to appeal to the fact that an even integer is of the form 2k for some integer k. In this case, since you START with a number that's defined to be even, it's enough to appeal to the fact that consecutive even integers from that number will have a difference of two.

@inkyvoyd, you're method is good, but the problem defines the variable B= the highest integer, so I don't see any point in REdefining it. Just use:
B +(B-2) +(B-4) +(B-6) +(B-8) = -20

Answer 15

I did. NOte my second post
x (greatest of the five) x+(x-2)+(x-4)+(x-6)+(x-8)=-20 Solve for x

Answer 16

@SmoothMath

Answer 17

Haha no. I'm saying you don't have to REDEFINE a variable that the problem already defines.

Answer 18

It's just extra work and confuses the issue to have two variables named the same thing.
B = highest even integer
x = highest even integer
We don't need both of these.