AMC 10 Daily Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 23 Hard

Let \lfloor x \rfloor denote the greatest integer less than or equal to x. Find the number of positive integer n that satisfy 2023\left\lfloor n\sqrt{{{2022}^{2}}+1} \right\rfloor=n\left\lfloor 2023\sqrt{{{2022}^{2}}+1} \right\rfloor.

  • A.

    0

  • B.

    1011

  • C.

    2022

  • D.

    3033

  • E.

    4044

Answer:E

By 2023\sqrt{{{2022}^{2}}+1}-2023\times 2022=2023\times \frac{1}{\sqrt{{{2022}^{2}}+1}+2022}<{}1,

∴ 2023\times 2022<{}2023\sqrt{{{2022}^{2}}+1}<{}2023\times 2022+1,

∴ \left\lfloor 2023\sqrt{{{2022}^{2}}+1} \right\rfloor=2023\times 2022,

∴ n\times 2023\times 2022=2023\left\lfloor n\sqrt{{{2022}^{2}}+1} \right\rfloor,

∴ \left\lfloor n\sqrt{{{2022}^{2}}+1} \right\rfloor=2022n,

So to get 2022n\leqslant n\sqrt{{{2022}^{2}}+1}<{}2022n+1, or n<{}\sqrt{{{2022}^{2}}+1}+2022, n\leqslant 4044.

Related Problems
Problem 21 AMC 10 Daily Practice Round 2
Problem 22 AMC 10 Daily Practice Round 2
Problem 24 AMC 10 Daily Practice Round 2
Problem 25 AMC 10 Daily Practice Round 2
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