AMC 10 Daily Practice Round 1

Complete problem set with solutions and individual problem pages

Problem 23 Medium

How many ordered pairs of positive integers (x, y) satisfy the equation x^2 + 615 = 2^y?

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:B

Consider the given equation x^2 + 615 = 2^y. Taking the equation modulo 3, we get: x^2 \equiv 2^y \pmod{3}. Since 3 \nmid 2^y, it follows that 3 \nmid x^2, so: x^2 \equiv 1 \pmod{3}. This implies that 2^y \equiv 1 \pmod{3}, which means that y must be even. Let y = 2m, where m \in \mathbb{N}^*. Substituting into the original equation, we get: (2^m - x)(2^m + x) = 615 = 3 \cdot 5 \cdot 41. We need to find pairs (2^m - x, 2^m + x) that satisfy this factorization. Solving for 2^m, we get: \begin{cases} 2^m - x = 5 \\ 2^m + x = 123 \end{cases} Adding and subtracting these equations, we find: 2^m = 64 \implies m = 6 \quad \text{and} \quad x = 59. Thus, y = 2m = 12, giving the solution (x, y) = (59, 12).

Therefore, there is only one positive integer solution, so the answer is \boxed{\text{B}}.

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