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Thread: Logic Q

7th September 2005 22:22 #526Registered User
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Come on now!

7th September 2005 22:25 #527Registered User
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OK 1st hint:
Try to think of the prices in cents not in $.

7th September 2005 22:32 #528Guest
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Re: OK 1st hint:
Originally posted by AminaDZ
Try to think of the prices in cents not in $.

7th September 2005 22:37 #529Registered User
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Re: Re: OK 1st hint:
Originally posted by phylay
Originally posted by AminaDZ
Try to think of the prices in cents not in $.

8th September 2005 17:32 #530Registered User
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Phew! Dear rival!
If we try to think of the prices in cents not in $ then we'd have:
w + x + y + z = 711
and
wxyz = 711000000
From the example:
7.11 = 3 x 3 x 0.79
We get:
711000000 = 3 x 3 x 79 x 10^6 (A)
It is clear that:
The smallest price possible: 0.01
The largest price possible: 7.08
But from (A) the largest price has to be a multiple of 79!
This gives us a choice of 79,158,237,316.
71179=632
711158=553
711237=474
711316=395
711000000/79=9000000
711000000/158=4500000
711000000/237=3000000
711000000/316=2250000
OMG! Now I'm stuck!!
...not sure if this is correct but I tried to minimize x+y+z subject to xyz = 2^n1 x 3^n2 x 5^n3 by choosing x,y, and z as close together as possible. The only option that seems to work is 2250000 where: 2250000 = 2^4 x 3^2 x 5^6.
By rearranging these factors, the values x,y, and z that make up 395 are:
x = 2 x 3 x 5^2 = 150
y = 2^3 x 3 x 5 = 120
z = 5^3 = 125
And so:
w = $3.16
x = $1.50
y = $1.20
z = $1.25
What's the correct way though?!!!!!
OMG! This is torture!

8th September 2005 20:44 #531Registered User
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Not an easy one I know!
I wanted to keep you thinking forever
OK the problem was given by Professor Doug Brumbaugh, University of Central Florida and here is the answer as given by Gérard P. Michon, Ph.D. (© 2000  2005 Gérard Michon)
Additional note from Professor Jim Wilson, University of Georgia:
I interpret the "exactly" and "exact" to mean just that  not "to the nearest penny."
Let's begin by representing the problem with a pair of equations.
xyzw = 7.11
x+y+z+w = 7.11
If x,y,z,w represent the amounts of the four items in cents then we have
xyzw = 711000000
x+y+z+w = 711
What can we say so far?
The factors : 711 = (3)(3)(79)
If there is to be an exact solution then the product xyzw must be divisible by 3, 9, 79.
The exact sum of the 4 items is 7.11 so the cost of each item is less than $7.11.
Proof:
If x, y, z, w are the prices of the items expressed in (whole) cents, what we are told is that x+y+z+w = 711 and xyzw = 711000000 = (2^6)(3^2)(5^6)(79).
So, one (and only one) of the amounts, say x, is a multiple of 79. This is, of course, less than 9 times 79 (which is 711), and may thus only be 1,2,3,4,5,6, or 8 times 79 (7 times 79 is ruled out, because it does not divide 711000000). These 7 possibilities translate into the following equations:
1. x=79, y+z+w=632, yzw=9000000
2. x=158, y+z+w=553, yzw=4500000
3. x=237, y+z+w=474, yzw=3000000
4. x=316, y+z+w=395, yzw=2250000
5. x=395, y+z+w=316, yzw=1800000
6. x=474, y+z+w=237, yzw=1500000
7. x=632, y+z+w=79, yzw=1125000
Now, the product of 3 positive numbers of given sum is greatest when they are all equal, which means that the product yzw cannot exceed (y+z+w)^3/27. This rules out the last three of the above 7 cases.
In the first three cases, on the other hand, the sum y+z+w isn't a multiple of 5, so at least one of y,z,w (say w) isn't either. Therefore, the product yz must be a multiple of the sixth power of 5. Since neither y nor z can be large enough to be a multiple of the fourth power [625 is clearly too big a share of 711, leaving only 7 cents for two items in the first case and nothing at all in the other two] we must conclude that both y and z are nonzrero multiples of 125. The number w would thus be obtained by subtracting from one of three possible sums (632, 553, 474) some multiple of 125 (necessarily: 250, 375 or [in the first two cases] 500). This gives only 8 possible values for w (382, 257, 132, 303, 178, 53, 224, 99). Each is unacceptable because it has a prime factor other than 2 or 3.
Therefore, only the fourth case is not ruled out, so that x=$3.16.
x=316, y+z+w=395, yzw=2250000.
Since the sum y+z+w is a multiple of 5, and at least one of the terms is a multiple of 5, either only one is, or all are. The former option is ruled out since this would imply for the single multiple of 5 to be a multiple of 5^6=15625, which would, by itself, be much larger than the entire sum of 395. So, y,z, and w are all multiples of 5, and we may let y=5y', z=5z', w=5w', where y'+z'+w'=79 and y'z'w'=18000.
Now, these three new variables may not be all divisible by 5 (otherwise their sum would be too). It's not possible either to have a single one of them divisible by 5, because it would then have to be a multiple of 53=125 and exceed the whole sum of 79. Therefore, we must have a multiple of 25 (say y'=25y") and a multiple of 5 (say z'=5z"): 25y"+5z"+w'=79 and y"z"w'=144. It is then clear that y" can only be 1 or 2. If y" was 2, then we would have 5z"+w'=29 and z"w'=72, implying that z" is a solution of a(295a)=72 or 5(a^2)29a+72=0. However, this quadratic equation does not have any real solutions, because its discriminant is negative. Therefore, y" is equal to 1 and y=5(25y")=125=$1.25.
The whole thing thus boils down to solving 5z"+w'=54 and z"w'=144, which means that z" is solution of a(545a)=144 or 5(x^2)54x+144=0. Of the two solutions of this quadratic equation (a=6 and a=4.8), we may only keep the one which is an integer. Therefore z"=6, z'=30, z=150=$1.50.
Finally, w'=144/z=24 which means w=5w'=120=$1.20.
Therefore, the solution is unique (the order of the 4 items being irrelevant):
$3.16, $1.25, $1.50, $1.20.

8th September 2005 20:52 #532Guest
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It wasn't easy
My turn now but an easy one as usual:
Divide $108 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $108, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?