Where was the plane?

[Aussiebear]

Air Show Show-Off (June 2025 Challenge Question)





Mason's father, Salvatore, is a pilot and earlier today Salvatore flew a plane in formation at an air-show.https://cr4.globalspec.com/PostImages/202505/9_planes_C0EA2800-AA23-2A5A-DA6E0C3DD8DCE558.png
"Which plane did you fly in?" asked Mason.
Salvatore drew a diagram of 9 planes and said, "The number of planes to right of me, multiplied by the number of planes left of me, is 3 less than it would have been if my plane had been 3 places to the right of me."
How did Mason solve this problem?




I'm assuming Salvatore was in position 6?

[Paul_Hossler]

I get Plane 4

But it depends on how your interpet "Three places to the right of me"

I did not count the plane he was in so 3 places to the right of #1 would be #7

32059

[Aussiebear]

I have now also come up with Plane #4 but my reasoning is slightly different.
32061

[Aussiebear]

I have seen the following answers



0x8=0, 1x7=7, 2x6=12, 3x5=15, 4x4=16, 5z3=15, 6x2=12, 7x1=7, 8x0=0
He was 3rd from the left, (2x6) = (5x3) - 3


and this which I'm yet to work through



(9-x)(x-1)=(9-(x+3))(x+2)-3


Which should be right up your alley Paul.....

And for those of you who are still thinking about the solution.
https://cr4.globalspec.com/PostImages/202506/IMG_0011_2AD4CCBF-F068-157F-2688C31CE7C19953.jpeg

[Aussiebear]

Which got me thinking if I used the formula in post#4....

If X = 1 then it compiles to 9 & 12, which would satisfy the desired outcome. However if X were to equal a value between 2 and 9 the results are wildly inaccurate.

[Aussiebear]

I have just seen another "answer"


FULL AND PROPER STEP BY STEP ALGEBRA:
L = planes to the left
R = planes to the right
L+R=8
Set it up [algebra is fun]:
[L * R] +3 = [L+3] * [R-3] --> NOW SUB IN "8-L" WHEREVER YOU SEE "R":
L[8-L] + 3 = [L+3] * [8-L-3]
8L - L^2 + 3 = [L+3] * [5 - L]
8L - L^2 + 3 = 5L - L^2 +15 - 3L
8L - L^2 + 3 = 2L - L^2 + 15
8L + 3 = 2L + 15
6L = 12
L = 2
SO WE CONCLUDE THAT IN THE ACTUAL FORMATION HE WAS #3. 2*6=12. HAD HE BEEN #6, THE MATH IS 5*3=15
12+3=15

...which puts me clearly in the sheep class.:banghead:. Not only do I not follow it, the explanation of the position I believe is wrong, or is it?

[p45cal]

Plane #4
32062

[Aussiebear]

Thank you P45cal.

[Logit]

There are 9 planes in a row. Salvatore flew one of them.
If Salvatore is in position x, then:


Planes to the left = x

Planes to the right = 8 - x

So, his current product = $$x \cdot (8 - x)$$
Now, if he were 3 places to the right, he would be in position x + 3:


Left = x + 3

Right = 8 - (x + 3) = 5 - x

So that product would be:$$(x + 3)(5 - x)$$
And we’re told:

Original product is 3 less than what it would have been if he were 3 spots to the right.
Putting that all together:$$(x)(8 - x) = (x + 3)(5 - x) - 3$$


Left side: $$8x - x^2$$
Right side: $$(x + 3)(5 - x) = 5x + 15 - x^2 - 3x = -x^2 + 2x + 15$$ Then subtract 3: $$-x^2 + 2x + 12$$
Now set the two sides equal:
8x−x2=−x2+2x+128x - x^2 = -x^2 + 2x + 12
We can eliminate −x2-x^2 from both sides:
$$8x = 2x + 12$$ Subtract 2x2x: $$6x = 12$$ So… �� x = 2

Final Answer:Salvatore was flying the 3rd plane from the left (since positions start at 0, 1, 2...).

[Aussiebear]

Plane #4 or as some of you would like to describe it "third from the left".