Wednesday

June 29, 2016
Posted by **gina** on Thursday, October 16, 2008 at 10:54pm.

but it was difficult for farmer Joe to find the

perfect pumpkin.

• Every third pumpkin was too small.

• Every fourth pumpkin was too green.

• Every fifth pumpkin had a broken stem.

• Every sixth pumpkin had the wrong shape.

How many perfect pumpkins did farmer Joe find

in the pumpkin patch?

- logic -
**SraJMcGin**, Thursday, October 16, 2008 at 11:00pmIf 440 were too smal, 330 were too green, 264 had a broken stem and 220 had the wrong shape, it must be 66 were perfect?

Sra - logic -
**Anonymous**, Thursday, October 16, 2008 at 11:08pmYep, that's what I got also: 66 were perfect.

- logic -
**Reiny**, Friday, October 17, 2008 at 1:46amthe question is not that simple

Suppose you line up all the 1320 and number them, then take out every third one.

So all the numbers divisible by 3 would be taken out.

then all the numbers divisible by 4 would be taken out, and

all those numbers divisible by 5 would be taken out

But if a number is divisible by 3 AND by 4 then it is clearly divisible by 6, so we can forget about the wrong shaped ones, since they are already out

What about the numbered pumpkins that are divisible by BOTH 3 and 5, such as 45

the 45 was already eliminated since it was divisible by 3 !!!!

So I will do it using Venn diagrams.

Call d3, d4, and d5 the numbers divisible by 3,4 and 5 respectively

I will draw 3 intersecting circles so there is common intersection of all three, (divisible by 3x4x5 or 60)

and the intersection of 2 at a time (3x4, 3x5, and 4x5)

1320/60 = 22

so put 22 in the intersection of d3 d4 and d5

1320/15 = 88 but that includes the 22 from the centre

so put 66 into the intersection region of d3 and d5 but not including d4

1320/12 = 110 but that includes the 22 from the centre

so put 88 in the intersection of d3 and d4 but not including d5

1320/15 = 88 but that includes the 22 from the centre

so put 66 in the intersection of d3 and d5 but not including d4

1320/3 = 440, this has to go into circle d3, but it already contains 66, 22 and 88 from the intersections, leaving 264 in the non-intersecting area of d3

1320/4 = 330, this has to go into circle d4, but it already contains 88,22, and 44 from the intersections, leaving 176 in the non-intersecting area of d4

1320/5 = 264, this has to go into circle d5, but it already contains 66, 22 and 448 from the intersections, leaving 264 in the non-intersecting area of d5

Now adding up all the entries ONLY ONCE we find in our circles we get 792

leaving 1320-792 or

528 perfect pumpkins- logic -
**gina**, Friday, October 17, 2008 at 2:21pmI did the Venn diagram the way you said and I didn't get the numbers to add up to 792...I have a redundancy of 264 in the non intersecting sections of d3 and d5. and then you mentioned the intersection of d4 and d5 to have a value of 44 and then a value of 448. so can you check this again or give me the instructions again because I must have misunderstood you...your answer makes sense but I still do not fully understand how you came to that answer using divisible numbers....thanks so much for the help

- logic -