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dominic
Email server is mostly setup now, you may now begin receiving emails from the forum...

I'd still like to add some custom reacts to the forum, let me know if you all have any ideas.
dominic
To new visitors the CME trading thread can be found in the FMA General Discussion Board
dominic
Let’s see who can solve this one:
IMG_0259.webp
B
Beesolberg
Let our integers be (n,n+2,n+4).

We then need the condition that 2n - 3(n+4) =15.
Solving:
2n-3n-12=15
-n=27
n=-27

We can verify by 2(-27)=-54 and 3(-23)=-69 so -54-(-69)=15. So our three integers are -27,-25,-23
dominic
dominic
Outstanding…
Curious choice of n, n+2, and n+4

I used n+1,3,5 and got it wrong first try cause I tried to do too much in my head… and then I got -27, -25, -23 and it didn’t look right cause negative numbers and inequalities… anyway

Great stuff!!
B
Beesolberg
The choice for n,n+2,n+4 came from the definitions of even and odd as 2n, and 2n+1 respectively.

I think the inequality can be dropped since it says "exceed 3x the greatest by 15" and didn't include an "or more" qualifier.
dominic
Welcome all to the MCFAM forum, or MCForum, or the FMA forum or [insert alternate name]
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