Last 3 digits of a perfect cube.

Since the last three digits of $n^3$ is $929$ , $n^3\equiv 929 \pmod {1000}$ $n^3\equiv 29 \pmod {100}$ $n^3\equiv 9 \pmod {10}$ If a cube of number ends with $9$ , then the number must also end with $9$ . $n\equiv 9\pmod {10}$ $\implies n^3=(10x+9)^3=(10x)^3+3\cdot (10x)^2\cdot 9+3\cdot (10x)\cdot 9^2+9^3$ $\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =1000x^3+2700x^2+2430x+729$ Considering the last two digits of $n^3$ , $n^3\equiv 30x+29 \equiv 29 \pmod {100}$ $\; \; \; \; \; \; \; \; \; \; \; \implies x\equiv 10 \pmod{100}$ So the tens digit of $n$ must be $0$ . $n\equiv 9 \pmod{100}$ $\implies n^3=(100y+09)^3=(100y)^3+3\cdot (100y)^2\cdot 9+3\cdot (100y)\cdot 9^2+9^3$ $\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\;\; \; =1000000x^3+270000x^2+24300x+729$ Considering the last three digits of $n^3$ , $n^3\equiv 300y+729\equiv929 \pmod{1000}$ $\; \; \; \; \; \; \; \; \; \;\implies 300y\equiv 200 \pmod{1000}$ $\; \;\; \; \;\; \;\; \;\; \;\; \;\implies y\equiv 4 \; \; \; \;\; \;\pmod{1000}$ So the hundreds digits must be $4$ . Therefore, the least value of $n$ is $\boxed{409}$ .

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Verification: $409^3=68417\underline{929}$

The last three digits of $n^3$ is $929$ implies that $\begin{cases} n^3 \equiv 929 \text{ (mod 1000)} \\ n^3 \equiv 29 \text{ (mod 100)} \\ n^3 \equiv 9 \text{ (mod 10)} \end{cases}$

Let $n \equiv 10a + b$ , where $a$ and $b$ are non-negative integers and $b \le 9$ . Then by Chinese remainder theorem :

$\begin{aligned} n^3 & \equiv 9 \text{ (mod 10)} \\ 1000a^3 + 300a^2b + 30ab^2 + b^3 & \equiv 9 \text{ (mod 10)} \\ \implies b^3 & \equiv 9 \text{ (mod 10)} \end{aligned}$

For $0\le b \le 9$ , only $9^3 \equiv 9 \text{ (mod 10)}$ . Therefore $b=9$ and $n = 10a + 9$ and

$\begin{aligned} 1000a^3 + 2700a^2 + 2430a + 729 & \equiv 29 \text{ (mod 100)} \\ 30a + 29 & \equiv 29 \text{ (mod 100)} \\ 30a & \equiv 0 \text{ (mod 100)} \\ \implies a & \equiv 0 \text{ (mod 100)} \\ \implies 100a & \equiv 0 \text{ (mod 100)} \\ \implies n & \equiv 100 a + 9 \end{aligned}$

And

$\begin{aligned} n^3 & \equiv 929 \text{ (mod 1000)} \\ 10^6 a^3 + 27\cdot 10^4 a^2 + 24300 a + 729 & \equiv 929 \text{ (mod 1000)} \\ 300 a + 729 & \equiv 929 \text{ (mod 1000)} \\ 300 a & \equiv 200 \text{ (mod 1000)} \\ \implies a & \equiv 4 \\ \implies n & \equiv 100a + 9 \equiv \boxed{409} \end{aligned}$