\(5. \text{Prove that if } n \ge 4 \quad then \quad n , n+2,n+4 \text{ can't all be primes.}\)

Give different methods of solving this. This was a problem to the OPC 2. The OPC 2 has already ended. OPC 3 will be released soon.Calvin sir suggested me to do this.So Thanks to him.\(\ddot\smile\)

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## Comments

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TopNewestStatement 2 : - A prime number n when divided by any number d , such that d < n , we get a remainder of either 1 or 2 .

Restating the above statement: .... If

(n =Prime Dividend, q = Quotient, and d = Divisor ) , given 0<d<n

Now , for n>= 4 , let d=3

then q must be greater than or at least equals to 1 , ie. q>= 1

case 1: For q > 1

so we can see that ( n , n+2 and n+4 ) cannot be all primes for each a) , b) and c) since they are product of 3 and a positive integer

case 2 :For q = 1

.for n=4 , n+2 and n+4 are not primes , also for =5 , n+4 is not prime .

Hence we can assert that for n>=4 n, n+2, n+4 cannot be all Primes

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one of them will be divisible by 3. so they cant all be primes

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Not necessarily..

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Sorry That Was A typo.The

real statement should be one of them will be divisible by 3 so they cant all be primes

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Let n be prime, so n is odd.

Case1: n mod 5=1, then n+4 mod 5 =0, so n+4 is not prime. Case2: n mod 5=3, then n+2 mod 5 =0, so n+2 is not prime.

Complete!

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Check my reply

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