<p>20) If k equals a positive integer, which of the following must represent an even integer that is twice the value of an odd integer?</p>
<p>A) 2k
B) 2k +3
c) 2k +4
d) 4k + 1
e) 4k+2</p>
<p>Since this was the last question, I knew it couldn't be A, but that seems like the only logical answer. The answer is E. Please Show me how</p>
<p>MUST is the key word here. If k were odd, it would work, but if k were even, it wouldn’t work (because then 2k would be twice the value of an even integer, not odd). Same thing for C. </p>
<p>4k+2 is twice of something. That something is 2k + 1. No matter what, 2k + 1 would be odd (because 2 * anything = even, and even + odd = odd).</p>
<p>Hope that helped, and good luck on the test!</p>
<p>Thank you so much. Now it all makes much more sense. So if I’m ever in doubt with what to do on a problem, is it generally a good idea to factor out anything possible, and see if that helps? (that’s how you got 2(2k+1) right?)</p>
<p>I factored 2 out because the question asks for “an even integer that is twice the value of an odd integer,” so the choices divided by 2 = the (possible) odd integers.</p>
<p>Glad I could help. (:</p>
<p>An even number is a number divisible by 2 (a number that, when divided by 2, results in an integer). Mathematically, it can be expressed as 2k, where k is any integer (including positive integers of course, as in your problem). An even number can be expressed as such because 2k is obviously divisible by 2. 2k/2 = k, and we said k is an integer, so 2k was clearly divisible by 2.</p>
<p>So if 2k is even, 2k + 1 must be odd. 2k + 1 is not divisible by 2 because (2k + 1)/2 = k + 1/2, which means you get an integer + 1/2, and that is not an integer. That’s one way of looking at it. The other way of looking at it is that if 2k is even, then the only way you get another even integer is by adding 2, not 1. </p>
<p>Anyway, since 2k + 1 is odd, twice that is 2(2k + 1), which is 4K + 2.</p>