Yitwail wrote, "I have six digits... "
Do I look like a Doctor? Get a saw.
"8x0249 (need a clue for the 10,000s digit?)"
Naw, you did good.
RapRap:
3) 6 digits
5 possibilities
810249
830249
850249
860249
870249
Now, what would you guys say if I told you the answer I had was-
800,249?
1) 11 cassettes
2) Fish
potential combinations (12)

they are
a,b,c
1,2,12
1.3.11
1.4.10
1.5.9
1,6,8
2,3,10
2,4,9
2,5,8
2,6,7
3,4,8
3,5,7
4,5,6
permutations on order--3 days is 3!=6
so 6*12=72
a) there are 12 combinations of the daily fish catch (in no particular daily order)
b) there are 72 combinbations of the daily fish catch in order.
4) 6 people, 5 horses
5) 4 dingos, 2 ostriches
Mark:
FISH
Same as Rap unless zero is allowed. (No freekin way is zero allowed, scheesh) :wink:
EYES AND LEGS
5 humans, 6 horses
(5+6)*2=22, 5*2+6*4=34
Additional eyes = 30, additional legs = 40
5 dingoes, 10 ostriches (15 animals)
(5+10)*2=30, 5*4+10*2=40
There is no solution with twice as many dingoes as ostriches if the additional eyes and legs are solely accounted for by the dingoes and ostriches.
Am I missing something here?
Well to be truthful - yes! There is. What I failed to mention:
Each human has 2 eyes and each horse has 2 eyes, so there were a total of 11 horses and people.
Let P = number of people and H = number of horses, then
H+P = 11 and P = 11-H
There were a total of 34 legs. Horses have 4 legs, people have 2, so we can write
4H + 2P = 34
If we replace the value of people in the second equation with the value of people from the first equation we get:
4H + 2(11-H) = 34
We can solve the rest fairly quickly:
4H + 22 - 2H = 34
4H - 2H = 12
2H = 12
H = 6
Plugging the value of horses back into the first equation we get:
P = 11-H = 11-6 = 5
There are 5 people and 6 horses.
In the second part we have a bit more to worry about, but we are provided with additional information. Let P = the number of people, H = the number of horses, W = the total number of wildlife, O = the number of ostriches, and D = the number of dingoes.
2P + 2H + 2W = 52 (eyes)
We know there are 5 humans and 6 horses so we get:
2(5) + 2(6) + 2W = 52
10 + 12 + 2W = 52
2W = 30
W = 15
The total number of all wildlife is 15 animals.
The 5 humans, 6 horses, and 15 wildlife have a total of 74 legs. If we subtract 10 legs for the 5 humans we are left with 64 legs, and if we subtract 24 legs for the 6 horses we are left with 40 legs. We were told that there were twice as many 4 legged creatures as there were 2 legged creatures so we are led to believe that O+D = 15, D=2(O), therefore 3(O) = 15, O=5 and D=10.
However, with 5 ostriches and 10 dingoes we should have 50 legs, but we have only 40 legs left for all the wildlife. Obviously either the numbers are wrong or some clue to the puzzle has been left out (on purpose).
If we start with the total number of legs left, we get 2(O)+4D=40, D=2(O), therefore 2(o)+4(2(O)) = 2(O)+8(0) = 10(O) = 40 and O=4, D=8. This leads us to one conclusion, of the 15 animals, 4 were 2 legged ostriches, 8 were 4 legged dingoes, and the remaining 3 were 0 legged snakes.
Ok, ok, I didn't mention snakes in the original puzzle, but sometimes you have to think outside the box a bit. I suppose you could have claimed that there were 5 ostriches and 10 dingo's and that the dingo's were each missing a leg, but the puzzle did state "4 legged dingo's", didn't it?
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Find as many possible solutions as you can
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On which dates would there be no problems assigned
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How much money did she start out with
Mark wrote, "Try: Were you unscathed by the explosions in London?"
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