Radio Astronomy Systems Research Group

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====== Answer =====

Sandy has $384615 and Sue has $538461

Let //X// and //Y// be Sandy's and Sue's money. Let //Z// be the number of dollars Sandy's statement. \\

//X+Z = Y-Z// and //2(X-Z) = Y+Z// \\
//X = 5Z// \\ 
//Y = 7Z// \\ \\
  
Let A be the first digit in Sue's wealth, B be the remaining digits, and n be the number of digits \\

//Y = 10<sup>(n-1)</sup>A+B// \\
//X = 10B + A// \\

//7X = 5Y// \\
//7(10B+A) = 5(10<sup>(n-1)</sup>A+B)// \\
//(5*10<sup>(n-1)</sup>-7)A=65B// \\

Since 65B is divisible by 5 and 5*10<sup>(n-1)</sup>-7 is not, A must equal 5 \\

//5*10<sup>(n-1)</sup>-7 = 13B// \\ \\
  
The least value of n such that 5*10^(n-1)-7 is a multiple of 13 is 6: \\

//499993=38461*13// \\

The next is 12, which makes them have more than 1 million dollars. Hence n=6, B=38461 \\

Radio Astronomy Systems Research Group

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