Question

Prove that if a positive integer is of the form 6 Q + 5 then it is the form 3 Cube + 2 for some integer cute but not conversely

Answer 1

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Let , 6q + 5 = x when q is a positive integer, 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 
x = 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 
x = 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 
x = 6q + 5 
= 6(3k + 2) + 5 
= 18k + 12 + 5 
= 18k + 17 
= 3(6k + 5) + 2 
= 3m + 2, where m is some integer 

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

Answer 2

Answer:

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

any

Positive integer is of the form 3k or 3k +1 or 3k + 2

There fore q= 3k or 3k + 1 or 3k + 2

If q = 3k then,

x= 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,

x= 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

x =6q+5

6(3k+2) +5

18k + 12 + 5

18k+ 17

3(6k+5) +2

3m +2 where m is some integer

Explanation:

If a positive integer is of the form 6q + 5 then it is of the form 3q + 2 for some integer q