SOLUTION: THIS WAS GIVIN TO ME BY MY TEACHER, AND IT'S JUST GOT ME STUMPED COMPLETELY. Clue 1: Safe combination... 15-22-?-5-38 Clue 2: Start with three consecutive integers between

Question 335536: THIS WAS GIVIN TO ME BY MY TEACHER, AND IT'S JUST GOT ME STUMPED COMPLETELY.
Clue 1:
Safe combination...
15-22-?-5-38
Clue 2:
Start with three consecutive integers between -10 and 10. twice the third is 13 greater than 3x the sum the first and the second.
Clue 3:
If you ever find those integers, multipy them together then divide them by -2.
Can you figure out the combination?
-Alexis Stephenson, Thank You.
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Since it states that we should "Start with three consecutive integers between -10 and 10. twice the third is 13 greater than 3x the sum the first and the second", this means that where 'x', 'x+1', and 'x+2' are the consecutive numbers.

Start with the given equation.

Distribute.

Combine like terms on the right side.

Subtract from both sides.

Subtract from both sides.

Combine like terms on the left side.

Combine like terms on the right side.

Divide both sides by to isolate .

Reduce.

Since , the three consecutive numbers are: -3, -2, -1

Multiply them out to get: (-3)*(-2)*(-1)=-6

Then divide that by -2 to get (-6)/(-2)=3

So the missing number is 3.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Let represent the least of the three consecutive integers. The second one is then and the next one is

Twice the third:

Three times the sum of the first and third plus 13:

So

Solve for , calculate the other two integers by adding 1 and then adding 1 again. Multiply the three together, then divide by -2. Let me know how you did.

John

My calculator said it, I believe it, that settles it

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