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Yes

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Let n be an integer. Then 2n is an even number

Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number.

Then:

even + odd = (2n) + (2m + 1)

= (2n + 2m) + 1

= 2(n + m) + 1

= 2k + 1 (where k = m + n) which is an odd number.

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