2020 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 24 Hard

How many positive integers n satisfy \frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor ?(Recall that [x] is the greatest integer less than or equal to \mathcal{x}.)(2020 AMC 10B, Question #24)

  • A.

    2

  • B.

    4

  • C.

    6

  • D.

    30

  • E.

    32

Answer:C

Solution 1:

First notice that the graphs of (x+1000) / 70 and \sqrt{n} intersect at 2 points. Then, notice that (n+1000) / 70 must be an integer. This means that \text{n} is congruent to 50(\bmod 70).

For the first intersection, testing the first few values of n (adding 70 to n each time and noticing the left side increases by 1 each time) yields n=20 and n=21. timating from the graph can narrow down the other cases, being n=47, \text{~d} n=50. This results in a total of 6 cases, for an answer of (\mathbf{C}) 6

Solution 2 (Graphing):

One intuitive approach to the question is graphing. Obviously, you should know what the graph of the square root function is, and if any function is floored (meaning it is taken to the greatest integer less than a value), a stair-like figure should appear. The other function is simply a line with a slope of 1 / 70. If you precisely draw out the two regions of the graph where the derivative of the square function nears the derivative of the linear function, you can now deduce that 3 values of intersection lay closer to the left side of the stair, and 3 values lay closer to the right side of the stair. With meticulous graphing, you can realize that the answer is (\text{C}) \text{G}. A in-depth graph with intersection points is linked below. https://www.desmos.com/calculator/e5wk9adbuk