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November 28, 2014

November 28, 2014

Posted by **Susan** on Monday, November 29, 2010 at 11:30am.

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.

- math -
**Henry**, Tuesday, November 30, 2010 at 12:10pmBad 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.

- math -
**Steve**, Friday, November 23, 2012 at 11:20amThat 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.

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