WAY = 100W + 10A + Y
rewrite the problem as
(10000W^2 + 2000WA + 200WY + 100A^2 + 20AY + Y^2)*1983 = 1000C + 100W + 10A + Y
Consider the last digit. Y^2*3 must end in Y. A little thought and the only possibilities are 0, 2, 5 and 7
Now consider the last 2 digits.
(20AY + Y^2)*83 must end in 10A+Y
Using each of the values above to find an A that works
If Y=0, A=0
If Y=2
(40A + 4)*83 = 100C + 10A + 2
3320A + 332 = 100C + 10A + 2
3310A = 100C  330
3310A = 100(C1) + 670
So A=7
Similar reasoning shows
If Y=5, A=7
If Y=7, A=4
So the number ends in 00, 72, 75, or 47
Now look at the last 3 digits
(200WY + 100A^2 + 20AY + Y^2)*983 must end in 100W+10A+Y
If A=0 and Y=0 then W=0 but this is not allowed
If A=7 and Y=2 then
(1400W + 4900 + 280 + 4)*983 = 1000C + 100W + 72
1376200W + 5095872 = 1000C + 100W + 72
1376100W = 1000C  5095800
1376100W = 1000(C10000)+4904200
So W = 2 So the number could be 272
Similar reasoning on the other endings finds 375 and 647

Posted by Jer
on 20090408 14:23:14 