math
posted by Susan .
Farmer Frank's pumpkin patch had 360 pumpkins and he was searching for perfect pumpkins to sell at his farm stand. As he walked through his pumpkin patch, he noted the following:
Every third pumpkin had no stem.
Every sixth pumpkin was too small.
Every fourth pumpkin was too big.
Every fifth pumpkin was not perfectly round.
How many perfect pumpkins were in Farmer Frank's pumpkin patch?
Only algebraic solutions will be accepted.

Bad Pumpkins = (1/3 + 1/6 + 1/4 + 1/5)360 = ((20 + 10 + 15 + 12) / 60)360 = (57/60)360 =(19/20)360 = 342.
Perfect P umpkins = 360  342 = 18. 
That can't be right. At least all the primenumbered pumpkins are perfect, and there are than 70 primes less than 360 (excluding 3 and 5).
I think you just didn't carry through the procedure far enough. If you scratch out all the 1/3 and 1/4, you have counted 1/12 twice, so you need to add it back in. Same for other pairs of factors.
360
360(1/3 + 1/4 + 1/5 + 1/6)
+360(1/12 + 1/15 + 1/18 + 1/20 + 1/24 + 1/30)
360(1/60 + 1/72 + 1/120)
+360(1/360)
= 124
But that doesn't count #1, which isn't one of the multiples. So, I think
125 is the final count.