When the square of any odd number, greater than 1, is divided by $8$, it always leaves remainder

### Concept & Formula This problem relies on the **Algebraic Properties of Odd Numbers**. Any odd integer can be universally represented by the algebraic expression $(2k + 1)$, where $k$ is any integer. ### Step-by-Step Solution * **Deduction via Algebra:** * Let an odd number be represented as $n = 2k + 1$. * The square of this odd number is: $$n^2 = (2k + 1)^2$$ * Expand the square using the identity $(a+b)^2 = a^2 + 2ab + b^2$: $$n^2 = 4k^2 + 4k + 1$$ * Factor out $4k$ from the first two terms: $$n^2 = 4k(k + 1) + 1$$ * **Calculation:** * Look at the term $k(k + 1)$. This represents the product of two consecutive integers. * The product of any two consecutive integers is **always even** (i.e., divisible by $2$). * Therefore, we can write $k(k + 1) = 2m$, for some integer $m$. * Substitute this back into our equation: $$n^2 = 4(2m) + 1$$ $$n^2 = 8m + 1$$ * This final form shows that $n^2$ is exactly $1$ more than a multiple of $8$. Thus, dividing by $8$ will always leave a remainder of $1$. ### Exam Strategy & Shortcut Do not use algebra in a timed exam for universal property questions. **Use Substitution.** Pick the smallest odd numbers greater than 1: Take $3$: $3^2 = 9$. $9 \div 8$ leaves remainder $1$. Take $5$: $5^2 = 25$. $25 \div 8$ leaves remainder $1$. Take $7$: $7^2 = 49$. $49 \div 8$ leaves remainder $1$. The pattern is instantly obvious. ### Common Pitfall Overthinking the algebraic proof can eat up valuable time. While the proof is mathematically elegant and important to understand, experimental verification with two simple test cases is vastly superior for competitive exams. ### Final Answer **Therefore, the correct answer is 1.**