Megan and Shana race against each other with the winner of each race receiving x
gold coins and the loser receiving y gold coins. (There are no ties and x and y are
integers with x > y > 0.) After several races, Megan has 42 coins and Shana has 35
coins. Shana has won exactly 2 races. The value of x is
(A) 3 (B) 7 (C) 5 (D) 6 (E) 4

1

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Suppose that Megan and Shana competed in exactly n races. Since Shana won exactly 2 races, then Megan won exactly n − 2 races. Since Shana won 2 races and lost n − 2 races, then she received 2x + (n − 2)y coins. Thus, 2x + (n − 2)y = 35. Since Megan won n − 2 races and lost 2 races, then she received (n − 2)x + 2y coins. Thus, (n − 2)x + 2y = 42. If we add these two equations, we obtain

(2x + (n − 2)y) + ((n − 2)x + 2y) = 35 + 42 or nx + ny = 77 or n(x + y) = 77.
 
Since n, x and y are positive integers, then n is a positive divisor of 77, so n = 1, 7, 11 or 77. Subtracting 2x + (n − 2)y = 35 from (n − 2)x + 2y = 42, we obtain ((n − 2)x + 2y) − (2x + (n − 2)y) = 42 − 35 or (n − 4)x + (4 − n)y = 7 or (n − 4)(x − y) = 7.

Since n, x and y are positive integers and x > y, then n−4 is a positive divisor of 7, so n−4 = 1 or n − 4 = 7, giving n = 5 or n = 11. Comparing the two lists, we determine that n must be 11. Thus, we have 11(x + y) = 77 or x + y = 7. Also, 7(x − y) = 7 so x − y = 1. Adding these last two equations, we obtain (x + y) + (x − y) = 7 + 1 or 2x = 8, and so x = 4. (Checking, if x = 4, then y = 3. Since n = 11, then Megan won 9 races and Shana won 2 races.

Megan should receive 9(4) + 2(3) = 42 coins and Shana should receive 2(4) + 9(3) = 35 coins, which agrees with the given information.)

Answer: (E)
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