Author Topic: Draft Lecture 1. Logical reasoning  (Read 29633 times)

yongcheng3315

  • Administrator
  • USAMO Winner
  • *****
  • Posts: 2585
Draft Lecture 1. Logical reasoning
« on: May 11, 2011, 06:40:50 PM »
Lecture 1. Logical reasoning (posted)

      File (1) Basic knowledge  (444.38 KB - downloaded 200 times.)
      File (2) problems and solutions (337.82 KB - downloaded 198 times.)
      File (3) Exercises and answer keys. (339.802 KB - downloaded 189 times.)

« Last Edit: October 27, 2011, 07:17:29 AM by A 2011 »

shanyecunmin

  • Math Coach
  • Mathcounts School Champion
  • ****
  • Posts: 3
Re: Draft Lecture 1. Logical reasoning
« Reply #1 on: May 14, 2011, 03:08:49 PM »
Thank you ! Very appreciate.

kellydai

  • Math Coach
  • Mathcounts School Champion
  • ****
  • Posts: 4
Re: Draft Lecture 1. Logical reasoning
« Reply #2 on: May 16, 2011, 12:29:38 PM »
thank you very much!

mathcount66

  • Math Coach
  • Mathcounts School Champion
  • ****
  • Posts: 5
Re: Draft Lecture 1. Logical reasoning
« Reply #3 on: May 16, 2011, 01:27:40 PM »
Thank you so much!!!

helenwyz

  • Math Coach
  • Mathcounts Chapter Champion
  • ****
  • Posts: 11
Re: Draft Lecture 1. Logical reasoning
« Reply #4 on: May 16, 2011, 11:23:37 PM »
Thanks a lot, Yongcheng.

聊天猫

  • 2012 AMC Summer Group
  • Mathcounts School Champion
  • *
  • Posts: 2
Re: Draft Lecture 1. Logical reasoning
« Reply #5 on: May 19, 2011, 11:06:22 PM »
Lecture 1. Logical reasoning (posted)

      File (1) Basic knowledge  (444.38 KB - downloaded 200 times.)
      File (2) problems and solutions (337.82 KB - downloaded 198 times.)
      File (3) Exercises and answer keys. (339.802 KB - downloaded 189 times.)

More chapters can be found here:

http://www.mymathcounts.com/Forum/index.php?board=43.0


Yongcheng,
I think problem 17 in the file (2) should be like this:
40-5=35
Every time the centipede goes up 5-3=2 feet
36/2=18
total is 18+1=19 days

yongcheng3315

  • Administrator
  • USAMO Winner
  • *****
  • Posts: 2585
Re: Draft Lecture 1. Logical reasoning
« Reply #6 on: May 20, 2011, 10:58:31 AM »
You are correct.

I made the correction several days back in the files posted in "50 lectures group" but forget to repost the new file here.

Thanks anyway.
« Last Edit: May 20, 2011, 11:19:57 AM by Andover 2011 »

xyz

  • Math Coach
  • Mathcounts Chapter Champion
  • ****
  • Posts: 15
Re: Draft Lecture 1. Logical reasoning
« Reply #7 on: May 22, 2011, 12:38:21 PM »
Chapter_1_Logical_reasoning_03  #7,   your answer is 42.

7. The teacher whispers positive integer A to Anna, B to Brett, and C to Chris. The students don’t know one another’s numbers but they do know that the sum of their numbers is 14. Anna says, “I know that Brett and Chris have different numbers” . Then Brett says, “I already knew that all three of our numbers were different” . Finally, Chris announces, “Now I know all three of our numbers”. What is the product ABC?

My son said that 42 is not the only answer, for example:if  A=1, B=2, C=11;
A+B+C=14   product ABC = 22.
if  A=1, B=5, C=8; A+B+C=14   product ABC =40. etc...
So, there are many answers. 
Is my son correct?

ppw

  • 2012 AMC Summer Group
  • Mathcounts School Champion
  • *
  • Posts: 2
Re: Draft Lecture 1. Logical reasoning
« Reply #8 on: May 23, 2011, 03:27:54 PM »
Anna says, “I know that Brett and Chris have different numbers” . ==> Anna has an odd number

Then Brett says, “I already knew that all three of our numbers were different” . ==> Brett has odd number 7, 9, or 11. Anna has 1, 3 or 5.  "ALREADY" means Brett did not use Anna's information.

Finally, Chris announces, “Now I know all three of our numbers”. ==> Chris has 6, Brett has 7 and Anna has 1.

===> 1X7X6 = 42
« Last Edit: May 23, 2011, 03:47:40 PM by ppw »

xyz

  • Math Coach
  • Mathcounts Chapter Champion
  • ****
  • Posts: 15
Re: Draft Lecture 1. Logical reasoning
« Reply #9 on: May 24, 2011, 08:32:19 PM »
But A can be 3, B can be 7, and C=4,
Then 3*7*4=84
or A=5  B=7  C=2   then 5*7*2=70,    etc....

ppw

  • 2012 AMC Summer Group
  • Mathcounts School Champion
  • *
  • Posts: 2
Re: Draft Lecture 1. Logical reasoning
« Reply #10 on: May 25, 2011, 09:56:01 AM »
C can not be 4. If C is 4 then (A, B) could be either (1, 9) or (3, 7). So Chris could not have known all three numbers.

C can not be 2 for the same reason. If C is 2 then (A, B) could be (1, 11), (3, 9), or (5, 7). So Chris could not have known all three numbers either.

If C=6, then the only possible (A, B) is (1, 7). So Chris could tell all three numbers.

Since C must be an even number less than 7, looks like 6 is the only answer.


jennifer

  • Math Coach
  • Mathcounts School Champion
  • ****
  • Posts: 6
Re: Draft Lecture 1. Logical reasoning
« Reply #11 on: May 27, 2011, 06:04:11 PM »
Here is my thoughts:
1). From Anna -> Anna is odd
2). From Brett -> Brett is odd, Chris is even, And Brett can only be 7, 9, or 11.
At this point, we still have 6 possibilities: 1+7+6, 3+7+4, 5+7+2, 1+9+4, 3+9+2, 1+11+2. Anna could be 1 or 3,  Brett could be 7,9, or 11, Chris could be 2,4, or 6
3). From Chris -> Since Chris says he knows all 3 numbers, then he must be 6. Because if he is 4, there are 2 possibilities. If he is 2, there are 3 possibilities. Only if he is 6, there is only one possibility which is: Anna must be 1 and Brett must be 7.

xyz

  • Math Coach
  • Mathcounts Chapter Champion
  • ****
  • Posts: 15
Re: Draft Lecture 1. Logical reasoning
« Reply #12 on: May 29, 2011, 12:14:40 PM »
When Anna says, “I know that Brett and Chris have different numbers” . Why  Anna must has an odd number? 

Brett can be 4, and  Chris can be 8, then Anna is 2.  Anna can has an even number, still satisfy A, B, C, all different numbers, and A+B+C=14.

yongcheng3315

  • Administrator
  • USAMO Winner
  • *****
  • Posts: 2585
Re: Draft Lecture 1. Logical reasoning
« Reply #13 on: May 29, 2011, 03:51:59 PM »
I see you are serious about math problem solving skills. I then really suggest that you register for the 50 Mathcounts lectures. Your child will learn a lot from it.

Good luck!

 


When Anna says, “I know that Brett and Chris have different numbers” . Why  Anna must has an odd number? 

Brett can be 4, and  Chris can be 8, then Anna is 2.  Anna can has an even number, still satisfy A, B, C, all different numbers, and A+B+C=14.

kevinyizhao

  • Math Coach
  • Mathcounts School Champion
  • ****
  • Posts: 3
Re: Draft Lecture 1. Logical reasoning
« Reply #14 on: July 05, 2011, 05:56:10 PM »
Why i can not see the draft for download? Do I have to register the class first? Where can I see some sample? Thx.