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Russian Mathematical Olympiad

Number theory Difficulty 5.0 AIME Prove it Russia

A 100-digit positive integer $n$ is called unusual if the last 100 digits of the decimal representation of $n^3$ form the decimal representation of $n$ , but the last 100 digits of the decimal representation of $n^2$ do not. Prove that there are at least two unusual 100-digit numbers. (V. Senderov)

Solution

For example, such numbers are $n_1 = 10^{100} - 1 = 99\ldots9$ and $n_2 = \frac{10^{100}}{2} - 1 = 49\ldots9$ . Indeed, the numbers
$n_1^3 - n_1 = (n_1 + 1)n_1(n_1 - 1) = 10^{100} \cdot n_1(n_1 - 1)$
and
$n_2^3 - n_2 = (n_2 + 1)n_2(n_2 - 1) = 10^{100} \cdot n_2 \cdot \frac{n_2 - 1}{2}$
are divisible by $10^{100}$ ; this means that $n_i^3$ ends with $n_i$ . On the other hand, the numbers $n_1^3 - n_1 = n_1(n_1 - 1)$ and $n_2^3 - n_2 = n_2(n_2 - 1)$ are not divisible by 5 (and even less by $10^{100}$ ); therefore, $n_i^2$ does not end with $n_i$ .

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