Question

How many of the first $1000$ positive integers can be expressed in the form $\lfloor2x\rfloor+\lfloor4x\rfloor+\lfloor6x\rfloor+\lfloor8x\rfloor,$ where $x$ is a real number, and $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z?$

Answer

Answer: 600

Concepts:floor and ceiling functions modular arithmetic counting integers in a range

Difficulty rating: 2340

Small Hint:

Substitute $y=2x$ and separate $y$ into its integer and fractional parts

Big Hint:

Determine the values of $\lfloor r\rfloor+\lfloor2r\rfloor+\lfloor3r\rfloor+\lfloor4r\rfloor$ for $0\leq r\lt1$

Solution:

Let $y=2x=m+r,$ where $m$ is an integer and $0\leq r\lt1.$ The expression is $10m+\lfloor r\rfloor+\lfloor2r\rfloor+\lfloor3r\rfloor+\lfloor4r\rfloor.$ Checking the breakpoints $r=\frac{1}{4},$ $\frac{1}{3},$ $\frac{1}{2},$ $\frac{2}{3},$ $\frac{3}{4}$ shows that the fractional-part contribution takes exactly the values $0,$ $1,$ $2,$ $4,$ $5,$ and $6.$ Thus precisely six residue classes modulo $10$ are attainable. Among the first $1000$ positive integers, this gives $100\cdot6=600.$

1985 AIME Problem 10

Answer: 600

Difficulty rating: 2340

Small Hint:

Big Hint:

Solution:

Problem 10 in Other Years