If RE*ER= AREA evaluate the AREA .
Please evaluate using pencil and paper only.
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^2100R10E1001A=0
which for a fixed value of A is actually a hyperbola rotated 45 degrees. Rearranging and factoring out R:
10R^2 + (101E100)R + (10E^210E1001A) = 0
Solving this quadratic for R involves the discriminant
(101E100)^2  4*10*(10E^210E1001A)
=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 = (100101Eħ(99E100))/20
R = 2E/20 = E/10 which is impossible
or R = (200200E)/20 = 1010E
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 20110408 15:16:18 