Show that an even integer is of the form 6q and 6q+2 for 6q+4 where q is a positive integer
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Let ‘a’ be any positive even integer and ‘b= 6’.
Therefore, a = 6q +r, where 0 ≤ r < 6.
Now, by placing r = 0, we get, a = 6q + 0 = 6q
By placing r = 1, we get, a = 6q +1
By placing, r = 2, we get, a = 6q + 2
By placing, r = 3, we get, a = 6q + 3
By placing, r = 4, we get, a = 6q + 4
By placing, r = 5, we get, a = 6q +5
Thus, a = 6q or, 6q +1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q +5.
But here, 6q +1, 6q + 3, 6q +5 are the odd integers.
Therefore, 6q or, 6q + 2 or, 6q + 4 are the forms of any positive even integers.
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Therefore, a = 6q +r, where 0 ≤ r < 6.
Now, by placing r = 0, we get, a = 6q + 0 = 6q
By placing r = 1, we get, a = 6q +1
By placing, r = 2, we get, a = 6q + 2
By placing, r = 3, we get, a = 6q + 3
By placing, r = 4, we get, a = 6q + 4
By placing, r = 5, we get, a = 6q +5
Thus, a = 6q or, 6q +1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q +5.
But here, 6q +1, 6q + 3, 6q +5 are the odd integers.
Therefore, 6q or, 6q + 2 or, 6q + 4 are the forms of any positive even integers.
I hope this helps you .
mark me as Brainlist .
❤️❤️
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