8. Let $n$ be a positive integer. By removing the last three digits of $n$, one gets the cube root of $n$. Find a possible value of $n$.
(1 mark)
Let $n$ be a positive integer. If the last three digits of $n$ are removed, the resulting number is the cube root of $n$. Find a possible value of $n$.
8. 32768
8. Let \( n = 1000a + b \), where \( 0 \leq b \leq 999 \). By removing the last three digits of \( n \), we get \( a \). Hence we have \( a^3 = 1000a + b \), or \( b = a(a^2 - 1000) \). Since \( b \geq 0 \), we have \( a^2 \geq 1000 \) and hence \( a \geq 32 \). If \( a \geq 33 \), then \( b \geq 33(33^2 - 1000) > 1000 \) which is impossible. Hence we must have \( a = 32 \), which gives \( b = 32(32^2 - 1000) = 768 \) and hence \( n = 32768 \).