<p>A list consists of 4 consecutive positive integers. Which of the following could be the number of integers in the list that are divisible by 13?</p>
<p>I. None
II. One
III. Two</p>
<p>I chose “only one” but apparently two is also possible. How can it be?</p>
<p>Thanks.</p>
<p>How would 2 be possible? I would got for I & II.</p>
<p>Yeah, I & II should be the answer. Where did you get this problem from? SAT prep books aren’t exactly error free.</p>
<p>Do you mean you chose “only I,” or you chose “only one such integer is possible”? Both I and II are possible. But III, that there are two multiples of 13 in a list of four consecutive integers, is impossible.</p>
<p>A list of four consecutive integers could contain one multiple of 13 (for example: 37, 38, 39, 40), but could also contain no multiple of 13 (for example: 7, 8, 9, 10).</p>
<p>Thanks all. I agree it is supposed to be I and II. There must be an error in the answer key. This question is from May 2011 QAS.</p>
<p>can someone please explain to me how 2 numbers in the list can be divisible by 13.</p>
<p>2 numbers in the list cannot be divisible by 13.</p>
<p>^^ Of course, but the OP is presenting a question that is not ETS like. Not at all. </p>
<p>A list consists of 4 consecutive positive integers. Which of the following could be the number of integers in the list that are divisible by 13?</p>
<p>Now, try the same problem but with this twist …A list consists of FOURTEEN oonsecutive positive integers. Which of the following could be the number of integers in the list that are divisible by 13? </p>
<p>I do not have that QAS, but that should be more like a real SAT question. The early one makes no sense with 4 integers.</p>
<p>xiggi, you are right! It was 14. My printed version was faulty :/</p>
<p>Well that changes everything:</p>
<p>The following list shows that III is possible:
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26 </p>
<p>The following list shows that II is possible:
12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25</p>
<p>And I is impossible because remainders are cyclical. If you start with a remainder of 1, then the 13th integer will have a remainder of 0:
1, 2, 3, 4, 5, 6,7 8, 9, 10, 11, 12, 0, 1</p>
<p>Oh, yeah, no small difference that!</p>
<p>Sent from my DROIDX using CC</p>
<p>Then it would be II and III.</p>
<p>This could be the future of the SAT. Here’s the question and here is the answer that does not make sense. Please tell us how to correct the stated problem to make the suggested answer plausible.</p>
<p>By the way, the difficulty of testing the divisibilty of intergers by 13 was a clear indication that the problem would be one of simple logic and not math. And that is why the problem pointed to a simple repetition in a 14 integers list. </p>
<p>ETS just wanted you to consider the possibility of a list that included the 13 and 26 - as in DSteve’s example.</p>