<p>I was doing this math problem, and I read the answer explanation. And like alot of the other explanations in test prep books, this one told me to just plug numbers in. I understand that part, but I think there must be a better (or just different) way to do it. here it is</p>
<p>g(x)= [(1/2)*x^2]/3</p>
<p>How many distinct integer values of g(x) are there that are less than or equal to 150 when x is a positive integer?
answer: 5</p>
<p>the book told me to plug every possible value in, which would get me 6, 24, 54, 96, 150… a total of 5 integers.
honestly, I know its really simple, but I spent so long trying to find a pattern in the first three so I wouldn’t have to plug in the rest. This is sort of for curiosity’s sake. Can anyone find a pattern or formula to use this, because I hate plugging in.</p>
<p>First solve for the exact value of x (integer or not) for which g(x)= 150. You have
g(x) = [0.5 x^2]/3 = 150
0.5 x^2 = 450
x^2= 900
x = 30</p>
<p>Also, for positive x values, g(x) increases as x increases - you will always have distinct values of g(x) for different positive x values, unlike say f(x) = sin(x). Finally, for g(x) to be integer, x needs to be a multiple of 3 (since g(x) = …/3) and a multiple of 2 (since g(x) = (1/2) * … ) i.e. a multiple of 6.</p>
<p>So, look for multiples of 6 in x = 1,2,3…, 30 . There are five such values: x=6,12,18,24,30 .</p>
<p>If plugging in the numbers is a possibility, it’s almost always the fastest and easiest way. It’s not as satisfying as actually figuring out the problem, but it’ll save you time on the test, and when you’re right, you KNOW you’re right.</p>
<p>thank you, it makes sense to work backwards.</p>