<p>(a-b)^n=(b-a)^n For all a and b, If n is an integer between 100 and 110 inclusive how many values of n are possible? </p>
<p>I'm not entirely sure about the wording of the question so if you remember it exactly please post it up. I must've thought about this question for a while without being able to decide between two answers: 6 and 11. The first option is the right answer if a does not equal b. In that case you need to have even numbers: 100, 102, 104, 106, 108 and 110. However the question did not state that a cannot equal b which is why I answered 11. If a=b you will get 0^n=0^n which is true for all n. This question cannot seem to be getting a definite answer. What do you guys think? 6 or 11?</p>
<p>“However the question did not state that a cannot equal b which is why I answered 11”</p>
<p>Umm, I don’t see the logic here. The equation is supposed to hold <em>for all</em> a and b. The correct answer is 6. For example, if n is odd, and a and b are different, then (a-b)^n and (b-a)^n are not equal, so the equation does not hold for all a and b if n is odd.</p>
<p>MITer correctly points out that the key is that the relationship must hold true for all values of a and b. </p>
<p>For example, if a=3 and b=3, and n=105, then yes the equation holds true. However, if a=5 and b=3, then the equation does not hold true for n=105. Therefore, n=105 does not work for all values of a and b. </p>
<p>In contrast, the even values of n will satisfy the equation irrespective of what values a and b take. </p>