Question

plz solve it faster.. I will mark as brainlist​

Answer 1

Step-by-step explanation:

Let, n = 6q + 5, when q is a positive integer

We know that any positive integer is of the form 3k, or 3k + 1, or 3k + 2

∴ q = 3k or 3k + 1, or 3k + 2 If q = 3k, then n = 6q + 5

= 6(3k) + 5

= 18k + 5

= 18k + 3 + 2

= 3(6k + 1) + 2

= 3m + 2, where m is some integer

If q = 3k + 1, then n = 6q + 5 = 6(3k + 1) + 5 = 18k + 6 + 5 = 18k + 11 = 3(6k + 3) + 2 = 3m + 2, where m is some integer

If q = 3k + 2, then n = 6q + 5

= 6(3k + 2) + 5

= 18k + 12 + 5

= 18k + 17

= 3(6k + 5) + 2

= 3m + 2, where m is some integer

Hence, if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q.

Conversely Let n = 3q + 2 We know that a positive integer can be of the form 6k + 1, 6k + 2, 6k + 3, 6k + 4 or 6k + 5

So, now if q = 6k + 1 then n = 3(6k + 1) + 2 = 18k + 5

= 6(3k) + 5

= 6m + 5, where m is some integer

So, now if q = 6k + 2

then n = 3(6k + 2) + 2

= 18k + 8

= 6 (3k + 1) + 2

= 6m + 2, where m is some integer

Now, this is not of the form 6m + 5

Hence, if n is of the form 3q + 2, then it necessarily won’t be of the form 6q + 5 always.

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