<p>This was not on the January 2010 SAT.</p>
<p>How many numbers are divisible by 3 between -3000 and 3000, inclusive?</p>
<p>oh god!
that reminds me of the question in Oct 09
where it was like</p>
<p>how many are not divisible by 3?</p>
<p>theres 333 numbers between 1 and 1000 divisible by 3
so, plain laziness and unwillingness to count + incompetence to reason with the question (lol) tells me that</p>
<p>there must also be 333 between 1001 and 2000, 2001 and 3000</p>
<p>so, 999 X 2(for the negatives) so 1998</p>
<p>i could be wrong, easily</p>
<p>3000 and -3000 themselves are divisible by 3 but all other endpoints are not</p>
<p>so if u include them, it would be 2000</p>
<p>which makes sense; every 3rd number is divisble</p>
<p>there are 3000 numbers each from 1->3000 and from -3000-> -1</p>
<p>so my final answer would be 2000</p>
<p>i could be wrong, again, easily</p>
<p>actually 2001 is also divisible by 3 but i dont think it shifts anything, and maybe the 3000s dont add to the total either…</p>
<p>all 2001 as an endpoint does is bump the next multiple of 3 to a 4 ones digit, then a 7 ones digit, and so on</p>
<p>3000 means the multiple before it ends in a 7 rather than a 6, as 999 could also serve as an endpoint</p>
<p>so now im just confused</p>
<p>i would say since there are 6001 numbers between -3000 and 3000, effectively there are 6000 because we know zzero is not divisible by 3(or is it?!?!), and following the every 3rd number thing, we can say that
there are 2000 numbers divisible by 3</p>
<p>although u shouldnt quote me on it</p>
<p>hope i helped?</p>
<p>The answer is 2001. There are 6001 numbers in the set [-3000,3000], an arbitrary 2000 of which are divisible by 3. The 6001st number, which is an end point, is also divisible by 3 because both end points (-3000 and 3000) are divisible by 3. Zero is divisible by all real numbers (besides 0). 0/3 = 0</p>
<p>bwahhh i just got owned by a property of zero</p>
<p>thanks lol</p>
<p>wow thanks guys~</p>