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 11-16-2012, 06:21 PM #1 New Member   Join Date: Oct 2011 Posts: 14 Daily Math Questions Hey guys, I'm prepping for the Dec SAT and I think a daily math questions would help me improve. Can anyone help me with these questions? If you nee help too, feel free to post. 1. Kyle's lock combination consists of 3 two-digit numbers. The combination satisfies the three conditions below. -One number is odd -One number is a multiple of 5 -One number is the day of the month of Kyle's birthday If each number satisfies exactly one of the conditions, which of the following could be the combination of the lock? A. 14-20-13 B. 14-25-13 C. 15-18-16 D. 20-15-20 E. 34-30-21 I'm confused because A, B, and D all fit the conditions... Reply
 11-16-2012, 06:35 PM #2 Member   Join Date: Jun 2012 Posts: 340 I'm not agreeing with how the question is given to us; but I'll give you the answer they're looking for. The only thing you missed was the word "one." ONE number is odd. ONE number is a multiple of 5. And ONE number.. you get it. A. 14-20-13 B. 14-25-13 - Eliminate because it has 2 odd numbers. C. 15-18-16 - Eliminate because 15 can't satisfy two requirements. D. 20-15-20 - Eliminate because it has two multiple of 5's. E. 34-30-21 - Eliminate because 21 and 30 work, but 34 doesn't satisfy the birthday requirement. A! Reply
 11-16-2012, 06:42 PM #3 New Member   Join Date: Oct 2011 Posts: 14 Omg I feel so stupid. Thank you so much! Here's another one I need help with: Let x be defined as [x]=x^2-x for all values of x. If [a]=a-2, what is the value of a? A. 1 B. 1/2 C. 3/2 D. 6/5 E. 3 Reply
 11-16-2012, 06:43 PM #4 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 This is not an SAT question, it's just an SAT-level question I made up. Q: If p and q are positive integers such that pq = 50, which of the following cannot equal (p^2)q? A: 50 B: 100 C: 200 D: 250 E: 500 Reply
 11-16-2012, 06:46 PM #5 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 @maraudersmap, are you sure your question is correctly typed? Here's what I did: [a] = a^2 - a = a - 2 a^2 - 2a + 2 = 0 a = (2 +- sqrt(4 - 8))/2, i.e. two complex solutions. Reply
 11-16-2012, 06:57 PM #6 Junior Member   Join Date: Apr 2011 Posts: 43 I'm pretty sure the question is: Let x be defined as [x]=x^2-x for all values of x. If [a]=[a-2], what is the value of a? Reply
 11-16-2012, 06:59 PM #7 New Member   Join Date: Oct 2011 Posts: 14 The question is typed correctly, [ ] is one of those weird symbols Reply
 11-16-2012, 07:11 PM #8 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 [a] = a-2 and [a] = [a-2] mean completely different things... Reply
 11-16-2012, 07:13 PM #9 New Member   Join Date: Oct 2011 Posts: 14 sorry, you guys are right. It's [a-2] Reply
 11-16-2012, 07:17 PM #10 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 Then it's just a^2 - a = (a-2)^2 - (a-2) a^2 - a = a^2 - 5a + 6 (upon expanding, simplifying). a^2 cancels, leaving -a = -5a + 6 --> a = 3/2 Reply
 11-16-2012, 07:21 PM #11 New Member   Join Date: Oct 2011 Posts: 14 Thank you very much! I'll post when I have more questions, I'm currently not at home haha [: Reply
 11-17-2012, 04:33 AM #12 New Member   Join Date: Jul 2012 Location: Austin, Texas Posts: 10 rspence, is the answer to your question C. ? Reply
 11-17-2012, 10:36 AM #13 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 ^JackTC yep. All the other answer choices are in the form 50p, where p is a factor of 50. Reply
 11-17-2012, 10:48 AM #14 Junior Member   Join Date: Sep 2012 Posts: 57 Can you go over how you got that question exactly? The way I did it took too long Reply
 11-17-2012, 10:54 AM #15 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 The solution involves thinking of (p^2)(q) as p(pq). Since pq = 50, this is equal to 50p. All the answer choices are multiples of 50, but note that p must be a factor of 50, so 50p must be (a factor of 50) times 50. 200 is 50*4, and 4 is not a factor of 50, so C is the answer. Reply

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