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# Prime numbers in A.P. | TOMATO Objective 152 If three prime numbers, all greater than $3$, are in A.P. , then their common difference

(A) must be divisible by $2$ but not necessarily by $3$;

(B) must be divisible by $3$ but not necessarily by $2$;

(C) must be divisible by both $2$ and $3$;

(D) need not be divisible by any of $2$ and $3$;

Discussion:

Say $p, p+d$ and $p+2d$ are the three primes in A.P. with d as the common difference.

Since $p>3$, hence $p$ is odd.

If $d$ is odd, then $p+d$ is even. But then $p+d$ cannot be a prime any more. Hence contradiction. Thus $d$ cannot be odd. Hence $2$ divides $d$.

Again $p> 3$ implies $p$ is not divisible by $3$. Hence p is either $1$ or $2$ modulo $3$. Now we wish to check if d is divisible by $3$ or not. Suppose it is not then d is also either $1$ or $2$ modulo $3$.

Case 1: $p$ is $1$ mod $3$ and $d$ is $1$ mod $3$, then $p+2d$ is $0$ mod $3$ hence contradiction (as p+2d is a prime greater than 3).

Case 2: $p$ is $1$ mod $3$ and $d$ is $2$ mod $3$, then $p+d$ is $0$ mod $3$ hence contradiction (as p+d is a prime greater than 3)

Case 1: $p$ is $2$ mod $3$ and $d$ is $1$ mod $3$, then $p+d$ is $0$ mod $3$ hence contradiction (as p+d is a prime greater than 3).

Case 2: $p$ is $2$ mod $3$ and $d$ is $2$ mod $3$, then $p+2d$ is $0$ mod $3$ hence contradiction (as p+2d is a prime greater than 3)

Hence $d$ must be $0$ modulo $3$.

Therefore common difference must be divisible by both $2$ and $3$.

Solving a few Diophantine Equations – Video

ISI 2015 BStat – BMath Objective Problems

If three prime numbers, all greater than $3$, are in A.P. , then their common difference

(A) must be divisible by $2$ but not necessarily by $3$;

(B) must be divisible by $3$ but not necessarily by $2$;

(C) must be divisible by both $2$ and $3$;

(D) need not be divisible by any of $2$ and $3$;

Discussion:

Say $p, p+d$ and $p+2d$ are the three primes in A.P. with d as the common difference.

Since $p>3$, hence $p$ is odd.

If $d$ is odd, then $p+d$ is even. But then $p+d$ cannot be a prime any more. Hence contradiction. Thus $d$ cannot be odd. Hence $2$ divides $d$.

Again $p> 3$ implies $p$ is not divisible by $3$. Hence p is either $1$ or $2$ modulo $3$. Now we wish to check if d is divisible by $3$ or not. Suppose it is not then d is also either $1$ or $2$ modulo $3$.

Case 1: $p$ is $1$ mod $3$ and $d$ is $1$ mod $3$, then $p+2d$ is $0$ mod $3$ hence contradiction (as p+2d is a prime greater than 3).

Case 2: $p$ is $1$ mod $3$ and $d$ is $2$ mod $3$, then $p+d$ is $0$ mod $3$ hence contradiction (as p+d is a prime greater than 3)

Case 1: $p$ is $2$ mod $3$ and $d$ is $1$ mod $3$, then $p+d$ is $0$ mod $3$ hence contradiction (as p+d is a prime greater than 3).

Case 2: $p$ is $2$ mod $3$ and $d$ is $2$ mod $3$, then $p+2d$ is $0$ mod $3$ hence contradiction (as p+2d is a prime greater than 3)

Hence $d$ must be $0$ modulo $3$.

Therefore common difference must be divisible by both $2$ and $3$.

Solving a few Diophantine Equations – Video

ISI 2015 BStat – BMath Objective Problems

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