My 11 year old brought home a typical 6th grade math challenge problem
and challenged me to solve it.

Here it is:

Show your work along with your answer...

The Apple Orchard:

Bishop's Apple Orchard harvested 18,000 apples this year.  Every third
apple harvested was too small (S), every fourth apple was too green
(G), and every tenth apple was bruised (B).  The remaining apples were
perfect (P). Here's a look at the first ten apples they harvested.

1st    2nd    3rd    4th    5th    6th    7th    8th    9th   10th   
P      P       S       G     P       S      P      G      S      B


How many of the 18,000 apples were "perfect"?

Hi peterstone1, 

First, I found the least common multiple of 3, 4 and 10 (since every
3rd, 4th and 10th apple was "bad"). The least common multiple is 60.

Knowing that, we now know that every 60 apples will have the same
number of "bad" and "perfect" apples.

1  P
2  P
3  S
4  G
5  P
6  S
7  P
8  G
9  S
10  B
11  P
12  S G 
13  P 
14  P
15  S
16  G
17  P
18  S
19  P 
20  G B
21  S
22  P
23  P
24  S G
25  P
26  P
27  S
28  G
29  P
30  S B
31  P
32  G
33  S
34  P
35  P
36  S G
37  P
38  P
39  S
40  G B
41  P
42  S
43  P
44  G
45  S
46  P
47  P
48  S G
49  P
50  B
51  S
52  G
53  P
54  S
55  P
56  G
57  S
58  P
59  P
60  S G B

Now after every 60 apples, the pattern will repeat itself, giving us
the same number of ?perfect? and ?bad? apples.

Going back and counting them, we know that in every 60 apples there will be:
20 apples that were too small (S) 
15 apples that were too green (G) 
6 apples that were too bruised (B)
28 apples that were perfect (P)

(Remember, sometimes apples were ?bad? in more than one way.)

Next, 18,000 divided by 60 is 300. So the above pattern will be counted 300 times. 

Which means of the 18,000 apples, there were:
300 x 20 = 6000 apples that were too small 
300 x 15 = 4500 apples that were too green 
300 x 6 = 1800 apples that were too bruised

and 

300 x 28 = 8400 apples that were perfect.


I hope that helps. If anything is unclear, please don?t hesitate to
ask for clarification. I will be glad to offer further assistance.

Best regards,
Rainbow

Well, you all got me to the right answer, despite a little confusion
in the middle.  I ended up wanting a quicker answer, and ran the
equation myself in excel to arrive at the 8,400 answer, so I knew the
right answer anyway, but it was interesting to see the interaction of
the various researchers.  Sucker5GA should probably avoid word/math
problems like this!

Here is how I would solve this. 

We have 18000 apples to start.

We need to find how many small apples are in the bunch so we clump
them into groups of 3.

18000 / 3 = 6000

so there are 6000 groups of 3 apples. For each group we can eliminate
1 as being too small.

OK?

Now we need to find apples that are too green so we group them into 4 apples.

18000 / 4 = 4500

and for each group of 4 there is one green apple.

Now we find bruised apples.

18000 / 10 = 1800.


So, we know there are 6000 small apples, 4500 green apples, and 1800
bruised apples
and we can subtract the total from 180000 to get the rest which are perfect. 

18000 - (6000 + 4500 + 1800) = 5700 perfect apples.


NOTE : I AM NOT A GOOGLE RESEARCHER, I TAKE NO RESPONSIBILITY FOR
INCORRECT ANSWERS!! USE AT YOUR
OWN RISK PLEASE!!!!!

Rainbow,
Good approach, very creative, however I think you are getting a
subtraction step wrong.
60 - 20 - 15 - 6 = 19     not   28

300 x 19 = 5700 apples.

No, it is Sucker who is wrong, and Rainbow correct. Sucker has
overlooked the fact that some apples are thirds, fourths and tenths,
and his method double counts those apples as non-perfect. You simply
have to count the number of Ps in Rainbow's first 60, to see that
there are 28, which are then repeated for each subsequent 60. I think
this is something of a "catch question" rather than just mathematics.

Hey geof,
I am a little confused about what you mean? I overlooked thirds...etc.
If you take this para from rainbows answer.

"Which means of the 18,000 apples, there were:
300 x 20 = 6000 apples that were too small 
300 x 15 = 4500 apples that were too green 
300 x 6 = 1800 apples that were too bruised"

there are a total of 12300 "bad" apples right?
Subtract 18000 - 12300 = 5700   ? 
right?
Am I just being dumb? It happens on occasion.

I get it, some of them overlap. good job. I stand corrected. Nice work rainbow.
sorry to contradict you.

You're making this too hard.  1/3 of the apples are too small, 1/4 are
too green and 1/10 are bruised.

1/3 of 18,000 = 6,000
1/4 of 18,000 = 4,500
1/10 of 18,000 = 1,800
total rejects = 6,000 + 4,500 + 1,800 = 12,300
18,000 - 12,300 = 5,700

Hi peterstone1, 

Sorry for the confusion in the middle. I couldn't find a better way to
explain it. :)
Thank you very much for the tip.

Best wishes,
Rainbow

Haha,
Thanks peterstone1, that's probably good advice to avoid such
problems. I havent done one since high school. bbescuela, sorry buddy
thats what I thought too but it  is wrong. Thats probably why rainbow
is a ga researcher and I am just a shmuck who pokes around on ga.:-)

good answer rainbow
sucker5 - some apples in his pattern of sixty are both small and 
Bruised
for example apple No. 30
there fore you cannot subtract the figures in the way that you have
because you are subtracting twice for that one apple.

if your interested in my way:

3 small= 6000
4 green= 4500
10 bruised= 1800


its all about lowest common multiples.

3x4x10 lcm = 60
3x4 lcm = 12
3x10 lcm = 30
4x10 lcm = 20

so in sixty,

apples which have all three defects in sixty = 60/60 = 1

apples which have two defects in sixty (s,g) = 60/12 = 5
apples which have two defects in sixty (s,b) = 60/30 = 2
apples which have two defects in sixty (g,b) = 60/20 = 3
 
now each of the above figures must be one less because one applies to
all three which should be counted twice (because its already been
counted once, but has three defects)

so in sixty 4, 1, 2, have two defects and 1 with all three defects to
be counted twice.


so 18000 - 
(
3 small= 6000
4 green= 4500
10 bruised= 1800

plus
duplications = 7(in 60) X 300 = 2100
treblications= 2(in 60) X 300 = 600
)




 6000
 4500
+1800
-----
12300

then subtract
 2100
  600
-----
 9600

then subtract 9600 from 18000 to get answer 8400.

I'm trying to do this problem using the inclusion-exclusion principle.
(if this problem didnt have such an small LCM this would be the most
practical way to approach it)

According to the IEP the number if good apples SHOULD be...

Total - |S|+|B|+|G|-(|SB|+|SG|+|GB|)+|SBG|
=
Total # of apples - ( Small + Bruised + Green - (Small&Bruised +
Small&Green + Green&Bruised)+ Small&Green&Bruised )
=
18000-(6000+1800+4500-(1500+450+600)+300)
=
7950


why don't our answers match?

|S|  =18000/3  = 6000
|G|  =18000/4  = 4500
|B|  =18000/10 = 1800
|SG| =18000/12 = 1500
|SB| =18000/30 = 600
|GB| =18000/40 = 450
|SGB|=18000/60 = 300

Nevermind..i found my problem.

|GB| should be 18000/20 not 18000/40  the least common multiple of 4
and 10 is 20 (not 40).

Despite my issues, though, I suggest that one uses the
Inclusion-Exclusion Principle to solve this type of problem rather
than trying to count apples (even if you only have a small subsection
to count).