CRV
how did you do it?
is there an sequences of equation that you used to do it?
Please explain
Thank you
how did you do it?
is there an sequences of equation that you used to do it?
Please explain
Thank you
I like the way you think Becks2k, let me try your equation again. I think that's the right way, but you probably made a mistake.
lets see
lets see
i'm assuming it
if stupid A+B=1.2038 and AB=1.2038 would just work out i had it
the other one works
if stupid A+B=1.2038 and AB=1.2038 would just work out i had it
the other one works
doesn't work
uh
ABCD=7.11
A+B+C+D=7.11
AB=E
A+B=E
CD=F
C+D=F
EF=7.11
E+F=7.11
E=1.203824
F=5.9062
CD=5.9062
C+D=5.9062
C=1.2754
D=4.63077
AB=1.2038
A+B=1.2038
doesn't work
uh
ABCD=7.11
A+B+C+D=7.11
AB=E
A+B=E
CD=F
C+D=F
EF=7.11
E+F=7.11
E=1.203824
F=5.9062
CD=5.9062
C+D=5.9062
C=1.2754
D=4.63077
AB=1.2038
A+B=1.2038
doesn't work
Yeah that's it...superhuman radioactive powers!
It's actually a well known puzzle that we had in Grade 8 and of course I couldn't remember all of it, but I did remember two of the numbers which made it alot easier.
It's actually a well known puzzle that we had in Grade 8 and of course I couldn't remember all of it, but I did remember two of the numbers which made it alot easier.
isn't this a system of equations problem, used w/ matricies? I'm too lazy to try it I remember doing something like this in 8th grade too... but just can't freakin remember... damn, that was only 5 yrs ago too...
I remember doing something like this in 8th grade too... but just can't freakin remember... damn, that was only 5 yrs ago too...It was quite a long time ago for me too..I do remember it took the class about two hours and 3 blackboards where covered with equations to try and explain it!
Right now I am eager to find the equations for this.
does anyone have a equation for this solution?
For CRV and others, you guys must have a really high 8th grade math class, cause I had algebra for it and still never anything like this before or at least related to it. I'm very impressed
does anyone have a equation for this solution?
For CRV and others, you guys must have a really high 8th grade math class, cause I had algebra for it and still never anything like this before or at least related to it. I'm very impressed
??
Dude i gotta be close
because 1.20*1.25=1.20+1.25
and i had A+B=A*B
and i got C+D=5.906
and the real answer 3.16+1.5+1.25=5.91
I think it has to be done somehow like I did it, just make one substitution mabye :/
Dude i gotta be close
because 1.20*1.25=1.20+1.25
and i had A+B=A*B
and i got C+D=5.906
and the real answer 3.16+1.5+1.25=5.91
I think it has to be done somehow like I did it, just make one substitution mabye :/
Hahah did a little search on the web, found this. Haha i just typed in,
-> math problem, 7.11
http://forum.swarthmore.edu/dr.math/problems/sam1.13.99.html
-> math problem, 7.11
http://forum.swarthmore.edu/dr.math/problems/sam1.13.99.html
uhh.... yeah.....
If you're expressing everything as cents, then why is ABCD = 711000000 or however many 0's there are...???
If you're expressing everything as cents, then why is ABCD = 711000000 or however many 0's there are...???
From that site, the solution.
Date: 01/14/99 at 11:35:42
From: Doctor Rob
Subject: Re: A+B+C+D=7.11 A*B*C*D=7.11
Thanks for writing to Ask Dr. Math!
This is fun, but a bit complicated.
Express all the amounts of money in cents, instead of dollars. Call the
amounts of money A, B, C, and D, which are all natural numbers. Then
A + B + C + D = 711
A*B*C*D = 711000000
1 <= A < 711
1 <= B < 711
1 <= C < 711
1 <= D < 711
First observe that the arithmetic mean of A, B, C, and D is
(A+B+C+D)/4 = 711/4 = 177.75
and their geometric mean is
(A*B*C*D)^(1/4) = 711000000^(1/4) = 163.29
It is a useful fact that the geometric mean (g.m.) of any set of
positive real numbers is always no greater than the arithmetic mean
(a.m.).
Now factor 711000000 into prime power factors
A*B*C*D = 2^6 * 3^2 * 5^6 * 79
One of them must be divisible by 79. Say it is A that is divisible by
79. Then A = 79*a, where 1 <= a < 9, and a | 2^6 * 3^2 * 5^6.
The fact that the arithmetic mean of B, C, and D is greater than their
geometric mean gives
(B+C+D)/3 = (711-79*a)/3 < (B*C*D)^(1/3) = [711000000/(79*a)]^(1/3)
493039*a^4 - 13312053*a^3 + 119808477*a^2 - 359425431*a + 243000000 < 0
This equation has solutions
0.94197... < a < 4.04065...
Because a is an integer, we have 1 <= a <= 4. Then
B + C + D = 711 - 79*a <= 632 < 3125 = 5^5
so 5^5 cannot divide B, C, or D.
Suppose first that B + C + D is not divisible by 5. Then exactly one of
them is not divisible by 5, and we label this one B. If 625 = 5^4
divides either C or D, say C, then 625 <= C <= 711 < 2*625, so C = 625.
Then B + D = 86 - 79*a >= 2, which implies a = 1, B + D = 7, and
B*D = 14400. Then the a.m. of B and D is 3.5 and their g.m. is 120,
which is impossible. Thus neither C nor D can be divisible by 5^4,
so both C and D must be divisible by 5^3 = 125. Write C = 125*c,
D = 125*d. Then we have B + 125*(c+d) + 79*a = 711, a*B*c*d = 9.
Then c + d < (711-79-1)/125, so 2 <= c + d <= 5. Also a, B, c, and d
can only be 1, 3, or 9, so one of c and d must be 1, say c, and then d
is 1 or 3. Now a*B*d = 9, B + 125*d + 79*a = 632. If d = 1, then
B + 79*a = 461 and a*B = 9, which lead to the quadratic equation
79*a^2 - 461*a + 9 = 0, which has no rational roots, a contradiction.
If d = 3, then B + 79*a = 211 and a*B = 3, which lead to the quadratic
equation 79*a^2 - 211*a + 3 = 0, which also has no rational roots, a
contradiction. So B + C + D not divisible by 5 is impossible.
Thus B + C + D must be divisible by 5, which implies that a = 4,
A = 4*79 = 316, B + C + D = 395, B*C*D = 2250000.
Now by an argument similar to the above, all three of B, C, and D must
be divisible by 5, so put B = 5*b, C = 5*c and D = 5*d, which gives
b + c + d = 79
b*c*d = 18000 = 2^4 * 3^2 * 5^3
As above, just two of b, c, and d are divisible by 5, and one of them
by 5^2. Say
b = 25*x, c = 5*y, 25*x + 5*y + d = 79, x*y*d = 2^4 * 3^2 = 144
Now x <= (79-5-1)/25, so 1 <= x <= 2. If x = 2, we get
5*y + d = 29, y*d = 72
and the a.m. of 5*y and d is 14.50, and their g.m. is 18.97, which is
impossible. Thus x = 1, and
5*y + d = 54, y*d = 144
This leads to the quadratic equation
5*y^2 - 54*y + 144 = 0
(5*y - 24)*(y - 6) = 0
y = 6 or y = 24/5
The last root is not an integer, so it is extraneous, and you get the
unique solution
y = 6
x = 1
d = 24
c = 30
b = 25
a = 4
A = 316
B = 125
C = 150
D = 120
and the amounts of money were $3.16, $1.50, $1.25, and $1.20.
- Doctor Rob, The Math Forum
http://forum.swarthmore.edu/dr.math/
Date: 01/14/99 at 11:35:42
From: Doctor Rob
Subject: Re: A+B+C+D=7.11 A*B*C*D=7.11
Thanks for writing to Ask Dr. Math!
This is fun, but a bit complicated.
Express all the amounts of money in cents, instead of dollars. Call the
amounts of money A, B, C, and D, which are all natural numbers. Then
A + B + C + D = 711
A*B*C*D = 711000000
1 <= A < 711
1 <= B < 711
1 <= C < 711
1 <= D < 711
First observe that the arithmetic mean of A, B, C, and D is
(A+B+C+D)/4 = 711/4 = 177.75
and their geometric mean is
(A*B*C*D)^(1/4) = 711000000^(1/4) = 163.29
It is a useful fact that the geometric mean (g.m.) of any set of
positive real numbers is always no greater than the arithmetic mean
(a.m.).
Now factor 711000000 into prime power factors
A*B*C*D = 2^6 * 3^2 * 5^6 * 79
One of them must be divisible by 79. Say it is A that is divisible by
79. Then A = 79*a, where 1 <= a < 9, and a | 2^6 * 3^2 * 5^6.
The fact that the arithmetic mean of B, C, and D is greater than their
geometric mean gives
(B+C+D)/3 = (711-79*a)/3 < (B*C*D)^(1/3) = [711000000/(79*a)]^(1/3)
493039*a^4 - 13312053*a^3 + 119808477*a^2 - 359425431*a + 243000000 < 0
This equation has solutions
0.94197... < a < 4.04065...
Because a is an integer, we have 1 <= a <= 4. Then
B + C + D = 711 - 79*a <= 632 < 3125 = 5^5
so 5^5 cannot divide B, C, or D.
Suppose first that B + C + D is not divisible by 5. Then exactly one of
them is not divisible by 5, and we label this one B. If 625 = 5^4
divides either C or D, say C, then 625 <= C <= 711 < 2*625, so C = 625.
Then B + D = 86 - 79*a >= 2, which implies a = 1, B + D = 7, and
B*D = 14400. Then the a.m. of B and D is 3.5 and their g.m. is 120,
which is impossible. Thus neither C nor D can be divisible by 5^4,
so both C and D must be divisible by 5^3 = 125. Write C = 125*c,
D = 125*d. Then we have B + 125*(c+d) + 79*a = 711, a*B*c*d = 9.
Then c + d < (711-79-1)/125, so 2 <= c + d <= 5. Also a, B, c, and d
can only be 1, 3, or 9, so one of c and d must be 1, say c, and then d
is 1 or 3. Now a*B*d = 9, B + 125*d + 79*a = 632. If d = 1, then
B + 79*a = 461 and a*B = 9, which lead to the quadratic equation
79*a^2 - 461*a + 9 = 0, which has no rational roots, a contradiction.
If d = 3, then B + 79*a = 211 and a*B = 3, which lead to the quadratic
equation 79*a^2 - 211*a + 3 = 0, which also has no rational roots, a
contradiction. So B + C + D not divisible by 5 is impossible.
Thus B + C + D must be divisible by 5, which implies that a = 4,
A = 4*79 = 316, B + C + D = 395, B*C*D = 2250000.
Now by an argument similar to the above, all three of B, C, and D must
be divisible by 5, so put B = 5*b, C = 5*c and D = 5*d, which gives
b + c + d = 79
b*c*d = 18000 = 2^4 * 3^2 * 5^3
As above, just two of b, c, and d are divisible by 5, and one of them
by 5^2. Say
b = 25*x, c = 5*y, 25*x + 5*y + d = 79, x*y*d = 2^4 * 3^2 = 144
Now x <= (79-5-1)/25, so 1 <= x <= 2. If x = 2, we get
5*y + d = 29, y*d = 72
and the a.m. of 5*y and d is 14.50, and their g.m. is 18.97, which is
impossible. Thus x = 1, and
5*y + d = 54, y*d = 144
This leads to the quadratic equation
5*y^2 - 54*y + 144 = 0
(5*y - 24)*(y - 6) = 0
y = 6 or y = 24/5
The last root is not an integer, so it is extraneous, and you get the
unique solution
y = 6
x = 1
d = 24
c = 30
b = 25
a = 4
A = 316
B = 125
C = 150
D = 120
and the amounts of money were $3.16, $1.50, $1.25, and $1.20.
- Doctor Rob, The Math Forum
http://forum.swarthmore.edu/dr.math/
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