I saw your reply re my late night solution to your pumpkin question
http://www.jiskha.com/display.cgi?id=1224212047
I did indeed make two typing errors, which even though they do not change the actual answer, nevertheless would have caused confusion.
I will repeat my solution and put the changed text into 'bold'
Here is the new solution:
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
1320/20 = 66 but that includes the 22 from the centre
so put 44 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 44 from the intersections, leaving 132 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
looks like I was "jumping lines" as I was copying from my scraps of paper solutions. Perhaps the fact that it was 1:40 in the morning might have had something to do with it, lol
sorry about that.
3 answers
first of all every 3rd would eliminate pumpkin #'s 3 6 9 12 15 18 21 24 ...
every 4th pumpkin would eliminate pumpkin
#'s 4 8 12 16 20 24 ....
notice that 12 and 24 appear in both runs
and those are multiples of 12 which is 3x4
so by dividing by 12 I would catch all those that appear in both runs, so that I don't count them twice in my Venn diagram.
I divided by 60 because 3x4x5 = 60 to catch all those that were in the elimination for 3 AND 4 AND 5
similarly 12 = 3x4
20 = 4x5 and
15 = 3x5 would catch all those that were in runs of (3 and 4), (4 and 5), and (3 and 5)
Does that make sense to you now?
Notice I also ignored multiples of 6. Why??