<p>Let S be the set of positive two digit integers in which the first digit is greater than the second digit. For example, 63 is a member of S, but neither 36 nor 33 is a member of S. How many integers are contained in S?</p>
<p>Hmm. I’m guessing what the question meant was that both digits are positive integers, which would eliminate 9 of the possibilities. If you typed the question in verbatim, then it is worded badly. I’m guessing it’s not from CollegeBoard.</p>
While zero IS an integer, its it NOT positive. Assuming this is correct, it would eliminate all the numbers that contain zero. Thus, 45 minus the 9 numbers that contain 0= 36.</p>
<p>Integer= positive or negative whole numbers…so yes, two digit integers exist. Two digit integers was included to specify that BOTH digits must be positive integers. And zero is not a positive integer.</p>
<p>As I read this thread I foresee the invention of a new math, or perhaps of a new definition of adjective in english – i.e. of how “positive” can modify each member of the number “two”.</p>
<p>First the answer to the original question is 45.</p>
<p>Second all the attempts to somehow squeeze a 36 out of 45 are sheer nonsense.</p>
<p>Third, the SAT is not a perverse exam. Twisted analysis is never required to solve a problem.</p>
<p>Actually fogcity, it’s just a case of ambiguous wording.</p>
<p>“Let S be the set of positive two digit integers” in the normal sense should give the answer 45. However, the makers of the question obviously didn’t intend this and worded it improperly, since their answer is 36.</p>
<p>By the way, College Board doesn’t make crap questions like this.</p>