24) Show that every prime number greater than 3 is of the form
6q+1 or 6q+5, where 'q' is some integer.
Answers
Given :- Show that every prime number greater than 3 is of the form 6q+1 or 6q+5, where 'q' is some integer.
Solution :-
we know that,
Euclid’s division Lemma:- States that any positive integer a can be divided by any other positive integer b in such a way that it leaves a remainder r .
- a = bq + r, where 0 ≤ r < b.
So,
Let a = 6q + r where 0 ≤ r < 6
if b = 6 , r can be 0,1,2,3,4 and 5 .
when r = 0,
→ a = 6q + 0
→ a = 6q
→ a = 2 * 3 * q
→ a = 2k where, k = 3q
Number is even. So, not a prime number.
when r = 1,
→ a = 6q + 1
Number is either product of primes or prime.
when r = 2,
→ a = 6q + 2
→ a = 2(3q + 1)
→ a = 2k, where k = (3q + 1)
Number is even. So, not a prime number.
when r = 3,
→ a = 6q + 3
→ a = 3(2q + 1)
→ a = 3k , where k = (2q + 1)
Number is divisible by 3. So, not a prime number.
when r = 4,
→ a = 6q + 4
→ a = 2(3q + 2)
→ a = 2k , where k = (3q + 2)
Number is even. So, not a prime number.
when r = 5,
→ a = 6q + 5
Number is either product of primes or prime.
Hence we can conclude that, every prime number greater than 3 is of the form 6q+1 or 6q+5, where 'q' is some integer.
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Answer:
Step-by-step explanation:
