Question

in how many pairs of consecutive prime number from 1 to 100, does a diffrence of 6 occur

Answer 1

Answer:

Required No. of pairs are = 7.

The prime Numbers from 1 to 100 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Pairs of consecutive prime numbers in which difference of 6 occurs:

So, required no. of pairs = 7.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

I’m assuming from the all steps prescription that we can’t just start with a list of primes.

To find what numbers of primes, we can discard all even numbers above 2, then consider their factors up to half the number in question.

(you can do the work here, do your own homework)

This should eventually give you your list of primes from 1–100:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

(Note 1 is NOT prime)

Now we can simply add 6 to each in turn and see if it equals the next number in the list, which gives us

5 + 6 is 11, but that’s not the next prime number, just a prime number.

(23, 29) works, as does (31, 37), (53, 59), (61, 67), (73, 79), (83, 89)

Incidentally if you’re wondering why this patterns turns up so much - of the next 6 numbers following a prime, 3 will be even and 2 will be divisible by 3 (with one being divisible by both), leaving only 2 possible non-primes one of which is always 6 higher than the previous prime.