Sandy and Sue each have a whole number of dollars. I ask them how many dollars they have.
Sandy says: “If Sue gives me some dollars, we’ll have the same amount of money. But if I give Sue the same number of dollars, she’ll have twice as much money as I have.”
Sue says: “And if you remove the first digit of my wealth and place it to the end, you’ll get Sandy’s wealth.”
If neither of them has more than 1 million dollars, how many dollars do they each have?
Hint
First get the ratio, then get their amounts.
Answer
Let X and Y be Sandy’s and Sue’s amounts of money, respectively. Let Z be the number of dollars in Sandy’s statement.
Then X+Z=Y-Z and 2(X-Z)=Y+Z. Solving this we get X=5Z and Y=7Z.
According to Sue’s statement, let A be the first digit of Sue’s wealth, B be the remaining digits, and n be the number of digits, then Y=A10^(n-1)+B and X=10B+A. 7X=5Y 7(10B+A)=5(A10^(n-1)+B)
(510^(n-1)-7)A=65B Since 65B is divisible by 5 and 510^(n-1)-7 is not, A must equal 5. We get:
510^(n-1)-7=13B The least value of n such that 510^(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 and B=38461.
Therefore, Sandy has 384615 dollars and Sue has 538461 dollars.
