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.
"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?
Last edited by Aussiebear; Yesterday at 03:34 AM.
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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
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Paul
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I have now also come up with Plane #4 but my reasoning is slightly different.
Screenshot 2025-06-25 at 04.39.32.jpg
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I have seen the following answers
and this which I'm yet to work through0x8=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
Which should be right up your alley Paul.....(9-x)(x-1)=(9-(x+3))(x+2)-3
And for those of you who are still thinking about the solution.
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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.
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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.. Not only do I not follow it, the explanation of the position I believe is wrong, or is it?
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Plane #4
2025-06-25_003711.jpg
Thank you P45cal.
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