SOLUTION: How would you work out this problem; It requires to show all work! The cafeteria has 360 roses to sell for Valentine's Day. You want to pick out a perfect red rose for someone

Question 269002: How would you work out this problem; It requires to show all work!
The cafeteria has 360 roses to sell for Valentine's Day. You want to pick
out a perfect red rose for someone special. When you get to the cafeteria
you discover:
*every third rose is too short
*every fourth rose has too many thorns
*every fifth rose is wilted
*every sixth rose is pink
*every eighth rose is yellow
If these are the only problems that the roses have, how many perfect roses are
there for you to choose from?
The problem requires to show all work. I would think that there is an
algebraic expression that would help me out, but I am in need of help.
I can help myself by making pictures to help visualize, but I think there must
be a better way.
Please help me if you can. Thank you!
Found 2 solutions by josmiceli, drk:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!

There are 120 roses that are too short

There are 90 roses that have too many thorns
But, since , every 12th rose was also divisible by 3,
and was also a too short rose,and
so 30 of the 90 are being counted twice. Just add
to the list of not wanted
So far I have not wanted

Of these every roses is being counted twice
and , so
66 roses are wilted and are not being counted twice
So far I have: not wanted
Every 6th rose is pink, but all these were counted as
a too short rose, since dividing by 6 is included by
dividing by 3, therefore don't count these
Likewise, every 8th rose was also counted when I counted
every 4th rose that had too many thorns,so don't count
these

114 roses of the 360 are perfect
I could have stumbled with my logic, but I
think this is right- hope yo at least un-
derstand what I'm doing

Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
It seems to me that you count 1, 2 and then primes greater than 5, and multiples of primes. Here is the list that I have:
1, 2,
7, 14, 49, 77, 91, 98, 119, . . . . . .. . . . . . .. [22 options]
11, 22, 121, 143, 154, . . . . . . . . . . . . . . [12 options
13, 26, 169, 182, 221, 247, 286, 299, 338 [9 options]
17, 34, 289, 323 . . . . . . . . . . . . . . . . . . . . .[4 options]
19, 38, 23, 46, 29, 58, 31, 62, 37, 74
41, 82, 43, 86, 47, 94, 53, 106, 59, 118,
61, 122, 67, 134, 71, 142, 73, 146, 77, 154,
79, 158, 83, 166, 89, 178, 91, 182, 97, 194,
101, 202, 103, 206, 107, 214, 109, 218, 113, 226,
127, 254, 131, 262, 137, 274, 139, 278, 149, 298,
151, 302, 157, 314, 163, 326, 167, 334, 173, 346,
179, 358,
----
181, 191, 193, 197, 199, 211, 223, 227, 229, 233,
239, 241, 251, 257, 263, 269, 271, 277, 281, 283,
293, 307, 311, 313, 317, 331, 337, 347, 349, 353,
359
----
so our total appears to be: 2 + 22 + 12 + 9 + 4 + 72 + 31 = 152 good options
---
On a personal note I didn't like this question. It could have been set up better using a smaller number such as 100 roses.
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