If the square of an odd natural number is divided by 8, then the remainder will be

## Concept & Logic Any odd natural number can be algebraically represented as $(2k + 1)$. When squared, its expanded form contains the product of two consecutive integers, forcing a specific divisibility property. ## Step-by-Step Solution * Let the odd natural number be $x = 2k + 1$, where $k$ is an integer ($k \ge 0$). * Square the number: $$(2k + 1)^2 = 4k^2 + 4k + 1$$ * Factor out $4k$ from the first two terms: $$4k(k + 1) + 1$$ * Observe the term $k(k + 1)$. This is the product of two consecutive integers. In any pair of consecutive integers, one must be even. Therefore, $k(k + 1)$ is always a multiple of 2 (i.e., $2m$). * Substitute $2m$ into the expression: $$4(2m) + 1 = 8m + 1$$ * This proves that dividing the square of an odd number by 8 will always exactly leave a remainder of 1. ## Exam Strategy & Shortcut Do not waste time on algebra if you can use pure arithmetic testing. Take the first few odd natural numbers, square them, and divide by 8: * $3^2 = 9 \Rightarrow 9 \div 8$ leaves remainder 1. * $5^2 = 25 \Rightarrow 25 \div 8$ leaves remainder 1. * $7^2 = 49 \Rightarrow 49 \div 8$ leaves remainder 1. The pattern is instantly clear and verifiable. ## Common Pitfall Students often test only the number 1 ($1^2 = 1$). While $1 \div 8$ mathematically leaves a remainder of 1, it can be confusing in the pressure of an exam. Testing 3 and 5 provides unambiguous confirmation. ## Final Answer **Therefore, the correct answer is 1.**