The number formed from the last two digits (ones and tens) of the expression $2^{12n} - 6^{4n}$, where $n$ is any positive integer is

### Concept & Formula The problem requires finding the last two digits of the algebraic expression $x^m - y^m$. We use the algebraic identity that $a^k - b^k$ is always perfectly divisible by $(a - b)$ for all positive integers $k$. If the difference $(a - b)$ ends in two zeros, the entire expression will end in two zeros. ### Step-by-Step Solution * **Given Expression:** $2^{12n} - 6^{4n}$ * Rewrite the bases to have a common exponent $n$: $$2^{12n} = (2^{12})^n = 4096^n$$ $$6^{4n} = (6^4)^n = 1296^n$$ * Now the expression is in the form of $a^n - b^n$, where $a = 4096$ and $b = 1296$. * According to the algebraic property, $a^n - b^n$ is divisible by $(a - b)$. * Calculate the difference: $$4096 - 1296 = 2800$$ * Since the base difference $2800$ is a multiple of $100$ (it ends in $00$), any integer power expansion or multiple of this base difference will also end in $00$. ### Exam Strategy & Shortcut Instead of using algebraic expansion, use the **Substitution Method**. Assume the simplest positive integer value for $n$, which is $n = 1$. Expression becomes $2^{12} - 6^4 = 4096 - 1296 = 2800$. The last two digits of $2800$ are $00$. This must hold true for all $n$. ### Common Pitfall Students often try to find the cyclicity of the last two digits of $2$ and $6$ separately, which becomes highly complex and time-consuming. Grouping the exponents to form a single power $n$ is the key. ### Final Answer Therefore, the correct answer is **00**.