Phil Ellinger
26/10/2013 05:23:49 am
144 = 2x2x(3!)x(3!)
Reply
George Lythgoe
26/10/2013 05:33:59 am
I got exactly half your answer: 72 = 2 x 3! x 3!
Reply
Phil Ellinger
26/10/2013 05:47:42 am
One number, 6, has to be at the intersection to be counted in both lines. The other three numbers in each line sum to 15. One of the two has the 8. The other two in that line can be 2+5 or 3+4. So there are two ways to partition the six numbers. The 8 can be in either the vertical or horizontal set. So 2x2x6x6.
Reply
George Lythgoe
26/10/2013 06:05:09 am
As the sum of all the numbers is 35, isn't it 7 at the intersection? This means the other three numbers need to sum to 14, leaving only one possible 'split' - 2,4,8 and 3,5,6.
Reply
Phil Ellinger
26/10/2013 06:20:43 am
You're quite right! I can cope with counting but not arithmetic! I probably shouldn't have tried to solve it in my head whilst watching Whitechapel! I must wait for next month's and give it my undivided attention...
Reply
Tom Thomson
26/10/2013 06:39:34 am
This is a lot easier than last month's puzzle.
Reply
Paul
26/10/2013 09:38:06 am
Thank you to Phil, George and Tom for your speedy responses.
Reply
George Lythgoe
26/10/2013 11:56:29 am
Just my 2p worth... I don't think it is too easy! It's accessible (so that anyone can have a go) but requires some careful thought to reach the correct answer. Enumeration is a really interesting area that seems to have all but disappeared from the A-Level syllabus. I think this puzzle would be an ideal introduction to the topic for students at this level.
Tom Thomson
26/10/2013 07:07:26 pm
I agree with George, it isn't too easy - just not as hard as last month's puzzle.
Sinh Trinh
26/10/2013 12:04:52 pm
The number of possible ways is 8. The proof is as follows.
Reply
Sinh Trinh
26/10/2013 03:09:22 pm
The answer is correct although there is one misstatement. Obviously, the arrangements are 8 hence, not unique. But the number at the intersection is unique which is what I meant to say.
Reply
Sinh Trinh
26/10/2013 06:06:37 pm
The correct answer is 12 different ways. There are 3! to arrange 3 different numbers in the vertical column and 3! to arrange 3 objects in the horizontal row.
Reply
Paul
27/10/2013 04:52:15 am
Thank you for your contribution Sinh. The answer won't be revealed for a while yet though.
Reply
Paul
27/10/2013 04:54:29 am
Thanks to George and Tom again for your helpful and interestingcontributions. I feel vindicated in choosing this puzzle now :)
Reply
mohammed ali abdulkarim
7/3/2014 02:00:34 am
vertical from top to bottom: 5,6,7,3
Reply
14/4/2014 12:38:57 pm
Comments say that last month's (Oct 2013) puzzle was easier. But I'm new to the site and only seeing puzzles from Nov 2013 - Apr 2014. Anyway I can see the older ones?
Reply
Trey
17/4/2014 11:46:21 am
I based this on even and odd numbers
Reply
## Leave a Reply. |
## Puzzle Ideas
If you have an idea for puzzle of the month, then please do let me know. ## Archives
April 2017
Don't forget to check out 'Teach Further Maths' - PowerPoints for Teachers and Students of A-Level Further Maths ...and the brilliant NEW educational card game 'Maths Trumps' |