A problem offering a solution

Question: If a, a + 2 and a + 4 are prime numbers, what are the number of possible solutions for a? (1) one (2) two (3) three (4) more than three.


The question appeared in CAT 2003, Re-test. The question belonged to Number Theory, and it suited me best to use Modular Arithmetic in solving it. Believe me, before I started I had not thought of its use, but when I saw that three consecutive members of an AP with a common difference of 2 were divisible either by 2 or by 3, with the result that remainder had to be zero in such division, the thoughts materialized themselves into giving a satisfactory solution. Here is the solution:


Sol: The three numbers a, a + 2 and a + 4 are all either even numbers or odd numbers.


If they are even, not more than one of them is prime {i.e. if a = 2}.

If they are odd, we have to show that none of them is divisible by 3 {since the numbers are a, a+2, a+4, if none of them is divisible by 3, none of them is divisible any other number except by itself of by 1, thus rendering them to be prime numbers}.


Since we are dividing the given numbers by 3, the remainders shall be only 0, 1 or 2.


If, suppose for number ‘a’, the remainder is 0, means the number is divisible by 3 and number is 3 and 3 is a prime number (we shall not consider 6 or 9 or 12 or ,….., because all these numbers are not primes).


Hence if a=3, a+2 = 5, a+4 = 7, all three are primes and this (a=3) is one solution.


Let a>3. Again on dividing by 3, the remainders are 0, 1 or 2.


If remainder is 0, the number is composite.

If remainder is 1, it can be written as, a == 1 mod 3. Adding 2 both the sides, we have, a+2 == 3 mod 3. Means a+2 is divisible by 3 and it is a composite number. Hence there is no solution.

Finally let the remainder be 2. Then a == 2 mod 3. Adding 4 both the sides, a+4 == 6 mod 3 è a+4 == 0 mod 3. Means a+4 is divisible by 3. Hence, once again there is no solution.


Hence the only solution for ‘a’ such that a, a + 2 and a + 4 are prime numbers, is a=3. Hence (1) is the answer.

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