Raziman's number

If an integer is a perfect $N$ -th power for $1 \le N \le 10$ , then it must be a perfect $\text{lcm}\{1,2,\ldots,10\} = 2^3 \cdot 3^2 \cdot 5 \cdot 7 = 2520$ -th power.

The smallest such integer greater than $1 = 1^{2520}$ is $2^{2520}$ .

Clearly, $2^{2520} \equiv 0\pmod{8}$ . Since $\phi(125) = 100$ , we have $2^{2520} \equiv 2^{20} \equiv (2^{10})^2 \equiv (1024)^2 \equiv 24^2 \equiv 576 \equiv 76 \pmod{125}$ .

Since $2^{2530} \equiv 0 \pmod{8}$ and $2^{2530} \equiv 76 \pmod{125}$ , by the Chinese Remainder Theorem, $2^{2520} \equiv 576 \pmod{1000}$ .

Therefore, the last $3$ digits of $2^{2520}$ are $\boxed{576}$ .

This is simple. For an $N$ th power, it's power must be atleast $lcm(1,2,...,10) = 2520$ . So smallest such number is $2^{2520}$ . We use modulo congruences to evaluate $2^{2520} \equiv 576 \pmod{1000}$ . (I'm omitting this as it was tedious calculation)

Let us call this smallest number n. To be a perfect Nth power. $n=m_N^N$ Where m is an integer which will depend on the power N and N is an integer between 1 and 10. This means n must have all Nth roots i.e. $n^{\frac{1}{N}}$ must be an integer for all N. So the smallest n will have a power that is divisible by all N. The smallest power we can gather that is divisible by all N is $2520=2\cdot2\cdot2\cdot3\cdot3\cdot5\cdot7$ Now the base will have to be the smallest integer greater than 1 which is 2. So smallest n is $2^{2520}$ The last three digits of which are 576.

Let $r$ be Raziman's number. The smallest perfect Nth power greater than $1$ is $2^N$ , as $1^N<2^N<3^N$ . If $r$ is a perfect power, it is in the form of $2^k$ . Also, $k$ must divide all of $1\le N\le10$ and be as small as possible, i.e., $k=\text{lcm}(1,2,3,\dots,10)=2^33^2(5)(7)=2520$ . From here, I used exponentiation by squaring to get the last 3 digits of $2^{2520}\equiv\boxed{576}\pmod{1000}$ .