AMC 10 Daily Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 21 Hard

What is the number of positive integer values of n \leqslant 2025 such that

x+\lfloor\sqrt{x}\rfloor+\lfloor\sqrt[3]{x}\rfloor=n has no real solution for x?

  • A.

    54

  • B.

    55

  • C.

    56

  • D.

    57

  • E.

    58

Answer:C

Since the floor function only outputs integers and n is an integer, x must be an integer. Let f(x)=x+\lfloor\sqrt{x}\rfloor+\lfloor\sqrt[3]{x}\rfloor for integers x \geqslant 0 (to prevent the number under the square root from being negative). If x is incremented by 1, x will always increase. While \lfloor\sqrt{x}\rfloor or \lfloor\sqrt[3]{x}\rfloor may not increase, they will never decrease. Thus, f(x) is a strictly increasing function along the nonnegative integers. Clearly, f(0)=0 and f(1)=3. It should be checked if there is an integer a such that f(a)=2025 (if not, the integer a where f(a) is as closest to 2025 as possible). Clearly, f(2025)>2025 and f\left(44^2\right)=1936+44+12=1992. Since f(x) is strictly increasing, 44^2 \leqslant a<45^2 if a exists. In this range, \lfloor\sqrt{a}\rfloor can only be 44 and \lfloor\sqrt[3]{a}\rfloor can only be 12.

Then, a=2025-44-12=1969, which lies in the range. Hence, f(1969)=2025. By the strictly increasing behavior of f(x), the values f(1), f(2), f(3), \cdots, f(1969) are distinct and the only values of f that can be one of the first 2025 positive integers. Thus, exactly 1969 of those integers are covered by f, so the number of integers n not covered by f is 2025-1969=56.