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Dmom
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Feb2703, 12:04 PM (EST) 

"math problem no answer in sight"

I beleive this is a problem from Benjamin Banneker you must spend $100 and buy 100 animals Cows $5.oo each Goats $1.00 each Chickens .05 cents each You must buy some of each animal I have worked many anwsers that are 99 & 100 but not 100 & 100 I would love to know the anwser thanks 

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MrToad
guest

Feb2703, 03:01 PM (EST) 

1. "RE: math problem no answer in sight"
In response to message #0

The unique answer is: 19 cows = $95 1 goat = $1 80 chickens = $4 To see this, note that you have two equations in two unknowns: (1) A+B+C=100 (2) 5A+B+0.05C=100 where A = # of cows, B = # of goats, and C = # of chickens. If you multiply (2) by 20 and then subtract (1), you get: (3) 99A+19B=1900 Now, A, B, and C are positive integers (I believe that it's implicit in the question that one can't buy, for example, 3/4 of a goat  otherwise, there are infinitely many solutions). Note that 19 divides 1900 and 19B, and so it must divide 99A. But, 19 is prime and does not divide 99, so it divides A. On the other hand, A cannot be greater than 20, because, from (2), 5A+(something positive)=100. Therefore A=19. From (3), one sees then that B=1, and then, from (1), one gets C=80. 

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CSpeed0001
Member since Feb1903

Feb2703, 03:01 PM (EST) 

2. "RE: math problem no answer in sight"
In response to message #0

I was able to get 19 cows, 1 goat, and 80 chickens. From the information that we have, we can generate 2 equations each with three unknowns, x+y+z=100 5x+y+.05z=100 and three (actually six) inequalities. 0<x<100 0<y<100 0<z<100 (Saying that we can't have negative numbers, or numbers bigger than 100. Actually, in this problem, one statement forces the other to be true.) Using whatever methods we have to solve linear equations (I used matrix algebra and a really powerful calculator), we get three equations down to two equations: x=(19/80)z y=(99/80)z+100 From here we can see that z has to be 80, or some product(?) of it (i.e. 160, 240, 320, etc). This is true because all of our numbers have to be whole numbers (unless the animals are dead meat). Z also has to be less than 100, because we can't buy a negative number of animals. That means z=80. From there, we can solve the rest of the equations by plugging 80 in for z. 

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Ben
guest

Feb2703, 03:01 PM (EST) 

3. "RE: math problem no answer in sight"
In response to message #0

Assuming that you have to have a whole number of each animal (and I think that that is a fair assumption) then it is not possible. Since chickens are $0.05 or 5 cents each in order to get a whole number of dollars you must have a multiple of 20 of them. That is you must have either 20, 40, 60 or 80 of them. The actual problem boils down to 2 equations, using H for cows, G for goats and C for Chickens 1) Fro money 5H + G + 0.05C = 100 2) From Animal numbers H + G + C = 100 Taking each of the 4 values of C we can have in turn then For C=20 1) 5H + G + 1 = 100 which equals 5H + G = 99 2) H + G + 20 = 100 which equals H + G = 80 (1)  (2) Gives (5H + G)  (H + G) = 99  80 or (51)H + (11)G = 19 or 4H = 19 or H = 19/4 This is not a whole number therefore can not be a solution. Performing the same calculation for C=40, C=60 and C=80 all gives the same result H is not whole. Therefore the problem has no solution. 

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Ben
Member since Nov1802

Feb2803, 10:40 AM (EST) 

4. "RE: math problem no answer in sight"
In response to message #3

>Performing the same calculation for C=40, C=60 and C=80 all >gives the same result H is not whole. Therefore the problem >has no solution. Doh, I really got it wrong, sorry. I put it down to trying to do the calculations after a day numbing my brain into submission peforming DVT VBI tests. Anyway for the case C=80 1) 5H + G + 4 = 100 which equals 5H + G = 96 2) H + G + 80 = 100 which equals H + G = 20 (1)  (2) Gives (5H + G)  (H + G) = 96  20 or (51)H + (11)G = 76 or 4H = 76 or H = 76/4 = 19 Put H=19 and C=80 into H + G + C = 100 and you get 19 + G + 80 = 100 pr G = 1 Same as everyone else. In my stupor last night I made the mistake 96  20 = 86 Once More Doh 

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malaka
guest

Jan0704, 08:56 PM (EST) 

14. "RE: math problem no answer in sight"
In response to message #3

when you change C to 40 and so on, your equations change, so it doesnt yields H=19/4 (what you call the same result) for all the values of C c=40 h=38/4=19/2 c=60 h=57/4 c=80 h=76/4=19 which gives you with, 80 chickens, 19 cows and 1 goat 80*0.05=4 euros 19*5=95 euros 1*1=1 euros 80+19+1=100 animals 4+95+1=100 euros as somebody stated before, so it does have a solution


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Whymme
guest

Apr1203, 03:57 PM (EST) 

5. "RE: math problem no answer in sight"
In response to message #0

And yet another way to compute it: First, leave the goats out. They are just used to fill up the number of animals until you reach 100. Then we're left with: 5X+0.05Y = X+Y, where X = cows and Y = chicken 5X+0.05YXY = 0 > subtract X+Y from both sides 4X0.95Y = 0 80X19Y=0 > multiply by 20 It's easy to see that the only solution possible is where X=19 and Y=80, so that you get 80x19 = 19x80 Whymme


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SnowBoB
guest

Jun2503, 11:48 AM (EST) 

6. "RE: math problem no answer in sight"
In response to message #0

Do you have to buy each one of the animals? If you don't you could do 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,$100! yeah! Oh, but that would'nt be 100 animals. we'll, i can't do it all here but try my email peabody_booboos@yahoo.com 

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g. sambasivan
guest

Jun2703, 03:56 PM (EST) 

7. "RE: math problem no answer in sight"
In response to message #0

fairly simple solution 19 cows for 95$ 1 goat for 1 $ 80 chickens 4$ 100 animals and 100$ is there any alternate solution; i think not. sambasivan 

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Puppkid
guest

Dec1503, 09:14 AM (EST) 

8. "RE: math problem no answer in sight"
In response to message #7

I got it! You get: 80 chickens + 14 cows + 16 goats = 100 animals. I did it the long way....simple adding....no equation. Here: 1.00 (20 chickens) +10.00 (2 cows) ________ $11.00 (22) + 10.00 (2 cows) _______ $21.00 (24) + 1.00 (1 goat) _______ $22.00 (25) + 1.00 (20 chickens) _______ $23.00 (35) + 10.00 (2 cows) _______ $33.00 (37) + 3.00 (3 goats) _______ $36.00 (40) + 10.00 (2 cows) _______ $46.00 (42) + 1.00 (20 chickens) _______ $47.00 (62) + 10.00 (2 cows) _______ $57.00 (64) + 5.00 (5 goats) _______ $62.00 (69) + 10.00 (2 cows) _______ $73.00 (71) + 10.00 (2 cows) _______ $83.00 (73) + 1.00 (20 chickens) _______ $93.00 (93) + 7.00 (7 goats) _______ $100.00 (100 animals)


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Flip
guest

Dec1703, 02:40 PM (EST) 

9. "RE: math problem no answer in sight"
In response to message #8

Time for a new calculator, Puppkid: 80 chickens @ $0.05 = $4 14 cows @ $5 = $70 16 goats @ $1 = $16 $4 + $70 + $16 = $90  you're still $10 short... ... but anyway: why does everyone have to reprove the solution that has already been given (and is unique within the constraints of the problem)? I am particularly astounded by the post that declares the problem impossible after two other people have already posted a solution!! 

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CSpeed0001
guest

Dec2703, 11:00 PM (EST) 

11. "RE: math problem no answer in sight"
In response to message #9

Sometimes, it takes a while for what we post to actually make it to the webpage. All three of us posted on the same day, so we didn't see the other two posts when we posted. The guy who said that there wasn't a solution hadn't the earlier two posts. CS 

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gilberto
guest

Jan2504, 09:46 AM (EST) 

18. "RE: math problem no answer in sight"
In response to message #0

Dmom, The number of chickens should be integer(multiples of 20). You have 4 choices for the chickens: x= number of cows y= number of goats
20 chickens= $1.00 ($5)x + ($1)y = $99.00 x + y = 20 no integer solutions 40 chickens = $2.00 ($5)x + ($1)y = $98.00 x + y = 60 no integer solutions 60 chickens = $3.00 ($5)x + ($1)y = $97.00 x + y = 40 no integer solutions 80 chickens = $4.00 (5)x + ($1)y = $96.00 x + y = 20 x=19 and y=1 80 chickens, 19 cows and 1 goat is the solution. 

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cow problem
guest

Mar0307, 08:14 PM (EST) 

22. "RE: math problem no answer in sight"
In response to message #0

20 chickens = 1.00 40 = 2.00 60 = 3.00 80 = 4.00 100 =5.00 so why did you evean put thie here it's INPOSABLE!!!duah unless you could have 100 chickens and then spend over 100 dollors, don't waste youre time. 

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