Problem 136

[mdean]

Nuts. I seem to be undercounting solutions to this one. My logic seems to make sense, but I only come up with 21 values of n<100. I may have to write a program to brute force this set of values to see what I'm missing.

I'd ask if someone was willing to pm the solutions for n<100 if it weren't for that dancing red text up above.

Update: All right, think I figured out the flaw in my logic and found 4 cases I was missing. So I just need to modify my program to reflect the change.

[satyres]

Hi, Can someone Clarfiy the problem for me ?
I don't understand "x, y, and z, are consecutive terms of an arithmetic progression"
Are the difference between x,y,z always 3 ??
Thanks for you help !

[mdean]

Hi, Can someone Clarfiy the problem for me ?
I don't understand "x, y, and z, are consecutive terms of an arithmetic progression"
Are the difference between x,y,z always 3 ??
Thanks for you help !

It means $x-y=y-z$.

[satyres]

It means $x-y=y-z$.

Thanks for the clarification ! but still have a problem !
the upper limit is there a mathimatical solution to find the max loop !
thanks !

[travisoneill]

The description seems to be incorrect. There are two valid solutions where n = 20. The provided x=13, y=10, z=7, and x=5, y=2, z=-1. Note that the description specifies that N must be a positive integer but gives no such constraint for x, y, or z. I guess you could consider (5,2,-1) a trivial solution nor N=20 like (-13,-10,-7) and (-5,-2,1) but without directly specifying a constraint on x,y,z here it makes it difficult to see what counts as a "unique" solution.

[mdean]

In the very first line, it says "The positive integers,$x,y$, and $z$,...