What is the largest of five consecutive even integers if the sum of the first four of them is A?

Question

the answer is: (A+20)/4

I just don't understand how they got to that.

Answer 1

Let x = the smallest.

Then the next four are represented by x + 2, x+ 4, x+ 6, and x + 8 (count by 2 for even integers).

The sum of the first 4 is given by A.

So, x + (x + 2), + (x + 4) + (x + 6) = A

Or, 4x + 12 = A.

So, x = (A - 12) / 4

But the largest of 5 is x + 8. Since x = (A - 12)/4, so x + 8 = (A - 12)/4 + 8 = (A-12)/4 + 32/4 =

((A - 12) + 32)/4 = (A + 20)/4

Try making up a specific example to see if this works.

Like: 6, 8, 10, 12, 14 (first four add to 36--which is A in your problem). (36+20)/4 should equal the largest -- 14?!

Answer 2

Hello Hannah!

First note that all even numbers are multiples of 2. I can say my first number is 2x, where x is any given integer number.

Therefore, five consecutive numbers are

2x

2*(x+1) = 2x+2

2*(x+2) = 2x+4

2*(x+3) = 2x+6

2*(x+4) = 2x+8

The sum of the first four is 8x + 12 = A

A/4 = 2x + 3

The largest of the five is 2x + 8 =

(2x + 3) + 5

The largest of the five is then A/4 + 5, which is the same as:

(A + 20)/4