Guest Posted July 20, 2009 Report Share Posted July 20, 2009 Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this alphametic equation. Each of A, P and R is nonzero. AREA = (PI)*(R)2 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 20, 2009 Report Share Posted July 20, 2009 R can not be less than 4 as it will not make a 4 digit number otherwise when mult by PI A can not be more than 7 as 98*81 = 7938 (max values of PI and R²) R can not be 9 either because if R2 = 81, I must be same as A Only possible values of R are 4,6,7 and 8 So, if A is odd, the only possible value for R is 7 If A is odd, no set of R and I satisfies the conditions except when A is 1, R is 7 and I is 9 then AREA must be 17E1 and R2 = 49 There is no value of E for which this is integer So, A must be even (2,4 or 6) Sets of A R and I that satisfy the conditions are: A R I 2 7 8 2 8 3 2 4 7 2 6 7 4 7 6 4 6 1 4 6 9 4 8 6 6 4 1 6 7 4 6 8 4 6 8 9 of these the only possible solutions are: 4704 = 49*96 and 4864 = 64*76 since in the latter there is a repetition, there is only one soln: 4704 = 96 * (7)2 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 20, 2009 Report Share Posted July 20, 2009 4 7 0 4 = 9 6 * (7)^2 A R E A = P I * ®^2 Quote Link to comment Share on other sites More sharing options...
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Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this alphametic equation. Each of A, P and R is nonzero.
AREA = (PI)*(R)2
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