+ Post New Thread
Results 1 to 15 of 15
General Chat Thread, Solved: My Son Went To A Maths Master-Class Today... in General; While he was there he was set this problem: Write down six numbers [from 1 to 9] Starting from the ...
  1. #1

    DaveP's Avatar
    Join Date
    Oct 2006
    Location
    Can't talk now: The mother-ship is calling!
    Posts
    8,784
    Thank Post
    351
    Thanked 1,274 Times in 870 Posts
    Blog Entries
    4
    Rep Power
    1126

    Solved: My Son Went To A Maths Master-Class Today...

    While he was there he was set this problem:

    Write down six numbers [from 1 to 9]

    Starting from the left add consecutive numbers together in pairs. If the result is bigger than nine record the right-most digit.

    Continue to do this until there are no more numbers to add together.

    Record the last number when the addition is complete.

    A worked example:

    Starting numbers:

    1, 2, 3, 4, 5, 6

    1st stage:

    3, 5, 7, 9, 1

    2nd stage:

    8, 2, 6, 0

    3rd stage:

    0, 8, 4

    4th stage:

    8, 4

    5th stage:

    2

    Another worked example [for clarity]

    Starting numbers:

    4, 7, 9, 8, 2, 1

    1st stage:

    1, 6, 7, 0, 3

    2nd stage:

    7, 3, 7, 3

    3rd stage:

    0, 0, 0,

    4th stage:

    0, 0

    5th stage:

    0

    The problem: Be able to predict the answer [at the 5th stage] without having to go through the stages in between.

    The tutor called this 'Modular Maths' and he said that when you see the solution it is very simple. I can't see the pattern at the moment ['A' level maths is a long time ago for me. However my son is 11!]

    Does anyone know how to explain this before it drives me crazy?

    Thanks.
    Last edited by DaveP; 8th November 2011 at 10:40 PM. Reason: Correct thread content.

  2. #2

    SimpleSi's Avatar
    Join Date
    Jun 2005
    Location
    Lancashire
    Posts
    5,808
    Thank Post
    1,476
    Thanked 592 Times in 444 Posts
    Rep Power
    168
    If there is a lone number at the end of the line with no partner to add to record that number alone.
    AFAICS This doesn't match with your description of the problem as there would be no "lone" number if pairing 1+2, 2+3, 3+4 etc

    You only end of with "lone" numbers if the idea is to add 1+2, 3+4, 5+6 (e.g 3,7,1) then next round you end up with 3+7,1 (e.g 0,1)

    Unless I've missed something?
    Si

  3. Thanks to SimpleSi from:

    DaveP (8th November 2011)

  4. #3

    DaveP's Avatar
    Join Date
    Oct 2006
    Location
    Can't talk now: The mother-ship is calling!
    Posts
    8,784
    Thank Post
    351
    Thanked 1,274 Times in 870 Posts
    Blog Entries
    4
    Rep Power
    1126
    Quote Originally Posted by SimpleSi View Post
    AFAICS This doesn't match with your description of the problem as there would be no "lone" number if pairing 1+2, 2+3, 3+4 etc

    You only end of with "lone" numbers if the idea is to add 1+2, 3+4, 5+6 (e.g 3,7,1) then next round you end up with 3+7,1 (e.g 0,1)

    Unless I've missed something?
    Si
    No I think you are correct. I was explained to me in somewhat of a rush. Your interpretation is correct. Having said that it does not shed any light anew for me.

  5. #4

    SimpleSi's Avatar
    Join Date
    Jun 2005
    Location
    Lancashire
    Posts
    5,808
    Thank Post
    1,476
    Thanked 592 Times in 444 Posts
    Rep Power
    168
    Nor me
    (but at least hopefully we've got the question right

    Si

  6. #5

    SimpleSi's Avatar
    Join Date
    Jun 2005
    Location
    Lancashire
    Posts
    5,808
    Thank Post
    1,476
    Thanked 592 Times in 444 Posts
    Rep Power
    168
    Think I've got it!

    Assuming you can ignore the modulus thingy at all stages and just do a final modulus then 5th stage equals

    a + 5b + 10c + 10d + 5e + f

    therefore you can ignore 10c and 10d as they will always produce 0.

    so answer is a + 5b + 5e + f

    i.e example 1

    1 + 10 + 25 + 6 = 42 equiv to 2

    example 2

    4 +35 + 10 + 1 = 50 equiv to 0

    QED (or maybe just lucky and it won't work for all cases )

    Si

  7. Thanks to SimpleSi from:

    DaveP (8th November 2011)

  8. #6

    MK-2's Avatar
    Join Date
    Oct 2006
    Location
    Nottingham
    Posts
    3,237
    Thank Post
    149
    Thanked 581 Times in 307 Posts
    Blog Entries
    8
    Rep Power
    199
    Is the answer "a Blue Whale?"

  9. #7

    DaveP's Avatar
    Join Date
    Oct 2006
    Location
    Can't talk now: The mother-ship is calling!
    Posts
    8,784
    Thank Post
    351
    Thanked 1,274 Times in 870 Posts
    Blog Entries
    4
    Rep Power
    1126
    Quote Originally Posted by SimpleSi View Post
    Think I've got it!

    Assuming you can ignore the modulus thingy at all stages and just do a final modulus then 5th stage equals

    a + 5b + 10c + 10d + 5e + f

    therefore you can ignore 10c and 10d as they will always produce 0.

    so answer is a + 5b + 5e + f

    i.e example 1

    1 + 10 + 25 + 6 = 42 equiv to 2

    example 2

    4 +35 + 10 + 1 = 50 equiv to 0

    QED (or maybe just lucky and it won't work for all cases )

    Si
    Looks good from here. However I am still curious to know if there is a more formal proof of this. There seems to be a proof for other 'trivial' problems like this. I can't believe that doesn't already have someone's name attached to it as a theorem/proof/conundrum/...

    Either way thanks again. Good work!

  10. #8

    DaveP's Avatar
    Join Date
    Oct 2006
    Location
    Can't talk now: The mother-ship is calling!
    Posts
    8,784
    Thank Post
    351
    Thanked 1,274 Times in 870 Posts
    Blog Entries
    4
    Rep Power
    1126
    Got it!

    It is, in effect, an 'inverted' Pascal's triangle. The sum can be described, in general terms, by the Binomial theorem.

    I should have spotted that ages ago but as I said my Maths 'A' Level was ages ago.

  11. #9
    pooley's Avatar
    Join Date
    Sep 2005
    Location
    S Wales
    Posts
    1,129
    Thank Post
    77
    Thanked 118 Times in 99 Posts
    Rep Power
    66
    Where's the tumbleweed gif when you need it



  12. #10

    X-13's Avatar
    Join Date
    Jan 2011
    Location
    /dev/null
    Posts
    9,033
    Thank Post
    591
    Thanked 1,945 Times in 1,345 Posts
    Blog Entries
    19
    Rep Power
    813
    Quote Originally Posted by MK-2 View Post
    Is the answer "a Blue Whale?"
    ~Le Claxon~

    -10 points from the QI elves. They knew you were going to say that.

    Also, are you secretly Alan Davies?

  13. #11

    Join Date
    Dec 2007
    Location
    cumbria
    Posts
    182
    Thank Post
    7
    Thanked 43 Times in 39 Posts
    Rep Power
    25
    Err, was probably watching the new series of Mongrels ...

  14. #12


    Join Date
    May 2009
    Location
    UK
    Posts
    2,105
    Thank Post
    256
    Thanked 450 Times in 251 Posts
    Rep Power
    141
    I don't get it yet...

  15. #13

    mac_shinobi's Avatar
    Join Date
    Aug 2005
    Posts
    9,710
    Thank Post
    3,246
    Thanked 1,048 Times in 970 Posts
    Rep Power
    364
    I don't get it either - I was bored silly in maths A Level and that was some years ago

  16. #14

    DaveP's Avatar
    Join Date
    Oct 2006
    Location
    Can't talk now: The mother-ship is calling!
    Posts
    8,784
    Thank Post
    351
    Thanked 1,274 Times in 870 Posts
    Blog Entries
    4
    Rep Power
    1126
    Quote Originally Posted by Rydra View Post
    I don't get it yet...
    Quote Originally Posted by mac_shinobi View Post
    I don't get it either - I was bored silly in maths A Level and that was some years ago
    What is it about this that you don't get?

    The question or the answer?

    I ask because I thought I had understood what was required initially but, as pointed out by @SimpleSi my original explanation of the problem was not correct.

  17. #15


    Join Date
    May 2009
    Location
    UK
    Posts
    2,105
    Thank Post
    256
    Thanked 450 Times in 251 Posts
    Rep Power
    141
    Well I can reliably get the right answer, but I've utterly failed to get a working equation out of it, and I did GCSE Maths and Statistics, A level maths, physics and Chemistry, along with both Chemistry and Maths courses at degree level.

    @SimpleSi: has pretty much got it.
    If you run with Simple's explanation, the middle two numbers have no effect on the final outcome, so literally ignore them.

    The next pairing of numbers (the 2nd and 5th numbers in the chain) provide a modifier according to the difference. If the difference between these two numbers is odd, then you will have a modifier of 5, but will have a modifier of 0 if the difference is even.

    The final pair of numbers (1st and 6th) is a straight forward addition.

    The simplest explanation I can give:

    1st number + 6th Number + 5x(2nd + 5th) = XX, and take the right hand number as your answer.

    What I couldn't get was an equation that could remove the left hand number reliably.

SHARE:
+ Post New Thread

Similar Threads

  1. [Video] Lip-Synch Human Voices To Dog's Action/Responses [My Son Loved It!]
    By DaveP in forum Jokes/Interweb Things
    Replies: 1
    Last Post: 14th May 2013, 02:22 PM
  2. My boss went to BETT ...
    By leco in forum BETT 2014
    Replies: 22
    Last Post: 19th January 2009, 07:08 AM
  3. My sons new pcs again.....
    By edie209 in forum General Chat
    Replies: 2
    Last Post: 22nd December 2007, 11:19 AM
  4. Replies: 1
    Last Post: 12th September 2007, 01:18 PM
  5. Replies: 6
    Last Post: 6th September 2007, 12:53 PM

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •