Posted by gina on .
There were 1320 pumpkins in a pumpkin patch,
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,
If 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,
Yep, that's what I got also: 66 were perfect.

logic 
Reiny,
the 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 nonintersecting 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 nonintersecting 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 nonintersecting area of d5
Now adding up all the entries ONLY ONCE we find in our circles we get 792
leaving 1320792 or
528 perfect pumpkins 
logic 
gina,
I 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