The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers. What is the smallest integer?

Question

having trouble solving 

Answer 1

5 consecutive even integers:

 

2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n

 

The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers:

 

2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 50

4n + 2 = 6n -32

34 = 2n

n = 17

 

The smallest integer in our sequence is 2n = 2*17 = 34

 

Check:

2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 50

34 + 36 = 38 + 40 + 42 - 50

70 = 120 - 50

70 = 70

Check!

 

Answer 2

This is just like the other one...

 

First assign a variable to the smallest of the numbers.  Let's call it x.

 

Since consecutive even integers are 2 apart from each other, that means that the other integers are:

 

x + 2

x + 4

x + 6

x + 8

 

So take what they tell you, and write it out using these values:

 

"The sum of the TWO SMALLEST of five consecutive even integers" -- that means x + (x + 2), or 2x + 2.

 

"the sum of the other three integers" -- that means (x + 4) +(x + 6) + (x + 8), or 3x + 18.

 

 

So the first sum is 50 less than the second, so 2x + 2 = (3x + 18) - 50

 

Now we can solve that:

2x + 2 = (3x + 18) - 50

2x + 2 = 3x - 32

2x + 34 = 3x

34 = x

 

So the smallest integer is 34.