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.

3 answers

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 prime-numbered 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.
144. 2 ways to solve:
1st response:
Step 1: take out all the 3s, 4s and 5s
(360 - (360 / 3) - (360 / 4) - (360 / 5)) = 78
Step 2: add back in all the overlaps of 3 and 4, 3 and 5, and 4 and 5
(78 + (360 / (3 * 4)) + (360 / (3 * 5)) + (360 / (4 * 5))) = 150
Step 3: take out the overlaps of 3, 4 and 5 (which were just taken out)
(150 - (360 / (3 * 4 * 5))) = 144

2nd method:
Count all the remaining numbers in the set of 1-30:
1, 2, 7, 11, 13, 14, 17, 19, 22, 23, 26, 29 -- 12
Multiply that times the number of sets of 1-30 in 360 (i.e., 12)
12 * 12 = 144