The remainder obtained when any prime number greater than 6 is divided by 6 must be (Campus Recruitment, 2007)

### Concept & Strategy This question tests the general algebraic form of prime numbers. A fundamental theorem in number theory states that every prime number greater than $3$ can be expressed in the form $6k \pm 1$, where $k$ is a natural number. ### Step-by-Step Solution Let's understand why the $6k \pm 1$ rule works. Any integer can be categorized by its remainder when divided by $6$. The possible forms are: 1. $6k$ (Divisible by $6$) 2. $6k + 1$ 3. $6k + 2$ (Factors out to $2(3k+1)$, divisible by $2$) 4. $6k + 3$ (Factors out to $3(2k+1)$, divisible by $3$) 5. $6k + 4$ (Factors out to $2(3k+2)$, divisible by $2$) 6. $6k + 5$ (Can also be written as $6k - 1$) Because primes greater than $3$ cannot be divisible by $2$ or $3$, we must immediately eliminate the forms $6k, 6k+2, 6k+3$, and $6k+4$. The only remaining possible forms for a prime number greater than $3$ (and therefore greater than $6$) are $6k + 1$ and $6k + 5$. * If a prime is of the form $6k + 1$, dividing by $6$ leaves a remainder of $1$. * If a prime is of the form $6k + 5$, dividing by $6$ leaves a remainder of $5$. Therefore, the remainder must be either $1$ or $5$. ### Exam Strategy & Shortcut You don't need to prove the theorem during the exam. Simply pick two or three prime numbers greater than $6$ and divide them by $6$ to find the pattern experimentally. * Prime $7$: $7 \div 6 = 1$ with Remainder **1** * Prime $11$: $11 \div 6 = 1$ with Remainder **5** * Prime $13$: $13 \div 6 = 2$ with Remainder **1** The pattern $1$ and $5$ emerges immediately, pointing directly to option (c). ### Common Pitfall Some students try to memorize complex divisibility rules instead of just testing real numbers. If you forget the $6k \pm 1$ formula under pressure, simply testing $7$ and $11$ safely guarantees the correct answer without theoretical knowledge. ### Final Answer Therefore, the correct answer is either 1 or 5.