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Find the area (Posted on 2011-04-08) Difficulty: 3 of 5
If RE*ER= AREA evaluate the AREA .

Please evaluate using pencil and paper only.

See The Solution Submitted by Ady TZIDON    
Rating: 4.0000 (3 votes)

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Hyperbolic solution. Seems to work. | Comment 1 of 5
Rewrite with R,E,A as variables: (10R+E)(10E+R)=1000A+100R+10E+A then expand and simplify:

100ER+10R^2+10E^2+ER = 1001A+100R+10E
10R^2+101ER+10E^2-100R-10E-1001A=0

which for a fixed value of A is actually a hyperbola rotated 45 degrees.  Rearranging and factoring out R:

10R^2 + (101E-100)R + (10E^2-10E-1001A) = 0

Solving this quadratic for R involves the discriminant
(101E-100)^2 - 4*10*(10E^2-10E-1001A)
=10201E^2 - 20200E + 10000 - 400E^2 + 400E + 40040A
=9801E^2 - 19800E + 10000 + 40040A

This discriminant must be a perfect square otherwise R will not be rational.  For some X it factors to
(99E + X)^2
Where 2*99*X = -19800 so X=-100 and X^2=10000
leaving 40040A=0
so A=0.  [Uh oh.  This was going so well.]

Finishing the quadratic equation above
R = (100-101Eħ(99E-100))/20
R = -2E/20 = -E/10 which is impossible
or R = (200-200E)/20 = 10-10E

The only 1 digit solution to this is E=1 and R=0

This makes the original alphametric
01*10=0010
which I suppose works since 1*10=10

AREA =10

This solution doesn't seem right because of the leading zeros.

  Posted by Jer on 2011-04-08 15:16:18
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