Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - [Enigma] Small problems

  1. #1
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6

    [Enigma] Small problems

    Hi,

    Here are some problems you can try to solve. I think it doesn't require much knowledge, but mostly tricks.

    #1 : simple algebra
    Less than 12 years ago, a mathematician, Vincent Urri, told his niece :
    - It's curious ! I have in front of me 3 positive integers. Their product is equal to my age and their sum gives your age !
    The niece, though she's good at calculations, wasn't able to guess what the 3 numbers were.

    This year, Vincent Urri has just turned 100 years old. How old is his niece today ?



    #2 : a little of number theory
    1. What is the remainder in the division of by 7 ?
    2. What is the remainder in the division of by 11 ?
    3. What is the remainder in the division of (where there are 502 times the pattern 2008) by 37 ?



    #3 : simple geometry
    The Swedish flag [Enigma] Small problems-drapeau_suede.gif has a yellow cross of constant width in a blue background. The yellow area represents one third of the flag's area.
    If the flag is 1m long and 66cm wide, what is the width of the yellow stripes ?



    #4 : aŁg€bra
    Teddy left his 9-digit code of his safe box in his safe box !
    Luckily, he recalls that the code doesn't have any 0, that the numbers are all different and that :
    - the number formed by the first 2 digits is a multiple of 2
    - the number formed by the 2nd and the 3rd digits is a multiple of 3
    - the number formed by the 3rd and the 4th digits is a multiple of 4
    - and so on... until the number formed by the 8th and the 9th digits, multiple of 9.

    With this information, he finds out 2 possibilities. What are they ?




    Enjoy
    Last edited by Moo; July 25th 2008 at 10:08 AM. Reason: his, not her, thanks galactus
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    For #2 part 1. Find the remainder when 2008^2008 is divided by 7.

    The mod 7 residues of the powers of 2008 are either 1 or 6.

    2008^0=1(mod 7)
    2008^1=6(mod 7)
    2008^2=1(mod 7)
    2008^3=6(mod 7)
    2008^4=1(mod 7)

    and so on................

    That whittles it down.

    Since 2008/4=502, the remainder is 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    For the first one, is the niece now 26?.

    The problem says, "less than 12 years ago". So, when Vincent was 90, then the product is 2*9*5=90.

    2+9+5=16

    That was 10 years ago, so the niece is now 26 yrs old.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by galactus View Post
    For the first one, is the niece now 26?.

    The problem says, "less than 12 years ago". So, when Vincent was 90, then the product is 2*9*5=90.

    2+9+5=16

    That was 10 years ago, so the niece is now 26 yrs old.

    What about this combination at 90: 3*5*6?

    The product is 90 and the sum is 14. Therefore, she could be 24 after 10 years.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    That appears to work as well. I reckon we're looking for only one that works though.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by galactus View Post
    That appears to work as well.
    The statement "less than 12 years ago" could mean he was 96. I am assuming all three numbers are different.

    96 = 3*4*8 with a sum of 15
    96 = 2*6*8 with a sum of 16

    which would put her at 19 or 20 now.

    I must be overthinking this one.....
    Last edited by masters; July 25th 2008 at 10:02 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Maybe this is nothing, but does 'told HIS niece' and 'how old is HER niece now?' have anything to do with it?. There appears to be confusion about the uncle/aunt's gender.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by galactus View Post
    Maybe this is nothing, but does 'told HIS niece' and 'how old is HER niece now?' have anything to do with it?. There appears to be confusion about the uncle/aunt's gender.
    It's a mistake a French person would often make
    There's no trap in it.

    And there are 3 solutions

    The thing is, you have to take care of the information "The niece, though she's good at calculations, wasn't able to guess what the 3 numbers were."

    None of your solutions are correct, and I need a complete demonstration
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    OK, I thought maybe it was one of those red herring problems.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by Moo View Post
    Hi,



    #3 : simple geometry
    The Swedish flag Click image for larger version. 

Name:	drapeau_suede.gif 
Views:	145 
Size:	322 Bytes 
ID:	7284 has a yellow cross of constant width in a blue background. The yellow area represents one third of the flag's area.
    If the flag is 1m long and 66cm wide, what is the width of the yellow stripes ?



    Enjoy
    Looks like the the flag area is 1m X .66m = .66 square meters.

    The yellow cross is one-third of that which is .22 square meters.

    Let W = the width of the yellow cross

    1 * W = W square meters for the horizontal stripe

    .66 * W = .66W square meters for the vertical stripe

    Subtract the intersection (W *W= W square) which was counted twice.

    Then we have: W + .66W - W^2 = .22

    This can't be right. Too much work.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by masters View Post
    Looks like the the flag area is 1m X .66m = .66 square meters.

    The yellow cross is one-third of that which is .22 square meters.

    Let W = the width of the yellow cross

    1 * W = W square meters for the horizontal stripe

    .66 * W = .66W square meters for the vertical stripe

    Subtract the intersection (W *W= W square) which was counted twice.

    Then we have: W + .66W - W^2 = .22

    This can't be right. Too much work.
    Well, it's right lol. Is this too much work ?

    but i must admit this problem isn't very *interesting*

    now, do the others , especially review the 1st problem :s
    Follow Math Help Forum on Facebook and Google+

  12. #12
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by Moo View Post

    now, do the others , especially review the 1st problem :s
    The guy had to have been between 88 and 100 years old at the time he talked to his niece.

    Between 100 and 88, there are only 3 numbers that can be factored into 3 single digit factors:

    90 = 2*5*9
    90 = 3*5*6

    96 = 2*6*8
    96 = 3*4*8

    98 = 2*7*7

    But you say none of the combinations is the correct solution.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by masters View Post
    The guy had to have been between 88 and 100 years old at the time he talked to his niece.

    Between 100 and 88, there are only 3 numbers that can be factored into 3 single digit factors:

    90 = 2*5*9
    90 = 3*5*6

    96 = 2*6*8
    96 = 3*4*8

    98 = 2*7*7

    But you say none of the combinations is the correct solution.
    What about 1 as a factor ?

    Since the girl can't know what the numbers are, *major hint* you have to find the triplets whose product is equal and whose sum is equal
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by Moo View Post
    #4 : aŁg€bra
    Teddy left his 9-digit code of his safe box in his safe box !
    Luckily, he recalls that the code doesn't have any 0, that the numbers are all different and that :
    - the number formed by the first 2 digits is a multiple of 2
    - the number formed by the 2nd and the 3rd digits is a multiple of 3
    - the number formed by the 3rd and the 4th digits is a multiple of 4
    - and so on... until the number formed by the 8th and the 9th digits, multiple of 9.

    With this information, he finds out 2 possibilities. What are they ?
    *781 254 963* and *187 254 963*

    I solved the problem this way :

    If the 9 digits of the code are a_1, a_2, ... , a_9 then the number a_8a_9 is a multiple of 9. As a_7a_8 is a multiple of 8, a_8 has to be even hence a_8a_9 is one of the following numbers : 27, 45, 63, 81.

    If a_8a_9 = 27 then a_7a_8 is a multiple of 8 which last digit is 2 hence it is either 72 (impossible since a_9=7) either 32. With a_7a_8=32, one has necessarily a_6a_7=63. (only multiple of 7 which ends by 3 and has two digits) As a_4a_5 is a multiple of 5, a_5 is 0 (impossible) or 5. Since there is no multiple of 6 which starts by 5 and ends by a_6=3 the case a_8a_9=27 is impossible.

    If a_8a_9=45 then ...


    This is really tedious, is there a better method to find the answer ?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by flyingsquirrel View Post
    *781 254 963* and *187 254 963*

    I solved the problem this way :

    If the 9 digits of the code are a_1, a_2, ... , a_9 then the number a_8a_9 is a multiple of 9. As a_7a_8 is a multiple of 8, a_8 has to be even hence a_8a_9 is one of the following numbers : 27, 45, 63, 81.

    If a_8a_9 = 27 then a_7a_8 is a multiple of 8 which last digit is 2 hence it is either 72 (impossible since a_9=7) either 32. With a_7a_8=32, one has necessarily a_6a_7=63. (only multiple of 7 which ends by 3 and has two digits) As a_4a_5 is a multiple of 5, a_5 is 0 (impossible) or 5. Since there is no multiple of 6 which starts by 5 and ends by a_6=3 the case a_8a_9=27 is impossible.

    If a_8a_9=45 then ...


    This is really tedious, is there a better method to find the answer ?
    Unfortunately no
    These are the correct answers.
    Next time I promise, I'll post more interesting problems
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Enigma machine
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: November 27th 2011, 10:35 PM
  2. Norm vectorial space: an "enigma"
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 23rd 2011, 03:52 PM
  3. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  4. small change from small changes
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 30th 2009, 08:30 AM
  5. Is A small?
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: May 28th 2009, 02:39 AM

Search Tags


/mathhelpforum @mathhelpforum