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Problem 467

AMC 12 late, AIME early

Number theory Difficulty 4.8 Find the answer

Find the smallest natural number $n$ such that the number $n^{2}$ begins with 2019 (i.e., $n^{2}=2019 \ldots$ ).

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

3. Since $\sqrt{2019}$ is not a natural number $\text{(44 2 =1936440 )}$ and $n$ 1420 and $n<\sqrt{20200} \cdot 10<1430$ . Now by examining this interval for $n$ that satisfies the given conditions, we already find for $n=1421$ that $1421^{2}=2019241$ .

Therefore, the smallest such natural number $n$ is $n=1421$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.