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El Cheapo Number Game

  1. Jan 16, 2005 #1

    Gokul43201

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    Follow the sequence of tsteps listed below, if you have the patience, and a calculator

    #0. Think of a positive integer (the smaller, the better for you)

    #1. Add 30 to it

    #2. Multiply this by 7

    #3. Write this number in the form 10X + Y, where [itex]0 \leq Y \leq 9 [/itex] (eg : N = 345 => X = 34, Y = 5)

    #4. Find A = |X - 2Y|

    #5. If A > 9 then write A = 10X + Y and repeat #4, else go to #6.

    #6. If A is any of 0,1,2,3,4 then find B = A+10, else if A is one of 5,6,7,8,9 then B=A+3

    #7. Find the B'th letter of the alphabet (1=>A, 2=>B, 3=>C, ...)

    #8. Think of a country beginning with this letter, in the Eastern Hemisphere (Europe, Africa, Asia, Oceania/Australia)

    #9. Take the last letter in this country's name

    #10. If possible, pick the smallest positive number whose spelling begins with this letter (T=>2, F=>4, S=>6), else pick the number corresponding to the position of this letter in the alphabet (a=>1, B=>2, C=>3, D=>4, G=>7, etc.)

    #11. Add 4 to this number.

    #12. Multiply this by any whole number less than 10.

    #13. Write this number, D as 10X + Y, just like before

    #14. Find E = |3X - Y|, and as before, repeat until E < 10

    #15. Multiply E by anyinteger you choose.


    Your final answer is..... :: ZERO ? ::

    Debunk the magic (if it worked :rolleyes:).
     
    Last edited: Jan 16, 2005
  2. jcsd
  3. Jan 16, 2005 #2
    Answer:
    Well, it doesn't. #4 can result in a negative number, as in:
    10 (+30)
    40 (*7)
    280 (10X+Y)
    X=28, Y=0 (X-2Y)
    28 (10X+Y)
    X=2, Y=8 (X-2Y)
    -14
     
  4. Jan 16, 2005 #3

    Gokul43201

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    Thanks for pointing that out. I forgot to put in a modulus sign (absolute value).
     
  5. Jan 16, 2005 #4
    Aha, I got it.

    The key is that the number at step 2 is divisible by 7. If you take a number N and multiply it by 7 to get M, you first multiply the ten's digit by 7. Then if you have one's place 0 in N you know that |X - 2Y| for M will also be divisible by 7. If the one's place in N is 1 then you add 0 to M's X and 7 to M's Y, so |X-2Y| will still be divisible by 7. Continuing in this fashion until you get to the one's place of N being 9, where you add 6 to M's X and 3 to M's Y, you find that |X-2Y| will always be divisible by 7 if M is divisible by 7.
    So in the end you must get 0 or 7, which in either case comes to a J, for which you have Japan or Jordan, both ending in n. You then get 9, then 13. Checking every possible number to multiply 13 by, from 1 to 9, you find it always comes back to 0.
     
    Last edited: Jan 16, 2005
  6. Jan 16, 2005 #5

    Gokul43201

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    Okay, that's a fair debunking.

    Now how about trying to find something more elegant than the "checking every possible number" approach ? In other words, can you prove the conjectures for < the divisibility rules for 7 and 13 > that I've cooked up ? In fact can you do better and establish the existance of such divisibility rules for any number ?
     
  7. Jan 16, 2005 #6
    Well, I did prove it for 7. I'll think about 13 later.

    By the way, negative numbers all start with n ("negative three, negative four...") and there is no smallest negative number so technically it doesn't quite work.
     
    Last edited: Jan 16, 2005
  8. Jan 16, 2005 #7

    Gokul43201

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    Okay, let's settle this once and for all. Negative numbers are a figment of your imagination. They are a myth propagated by the "New Evil Geniuses About To Vaporize Everyone", a cartel of mathematical villains striving for global domination.
     
  9. Jan 17, 2005 #8
    Well, you can do 13 similarly to how I did 7.
    You start with some number written in decimal form AA...AB. To multiply it by 13 you first multiply the AA part by 13 and then the B (units) part by 13 and add them together. When B is 0, |3X - Y| will be divisible by 13 since it is just the A's multiplied by 13 * 3. When B is 1, it's the same as if B were 0, except that to 3X you must add 1 * 3 = 3, and to Y you must also add 3, so |3X - Y| comes back to the same value. When B is 2, it's the same as if B were 0, except that to 3X you must add 2 * 3 = 6, and to Y you must also add 6, so |3X - Y| comes back to the same value. And so on, until when B is 9, 13 * 9 = 117, so that to 3X you must add 11 * 3 = 33, and to Y you must add 7, so that |3X - Y| is 26 greater than what it would have been if B were 0, so it is still divisible by 13.
     
  10. Jan 17, 2005 #9

    Gokul43201

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    Here's another proof I came up with : it's for a general case, and so can be used to generate a divisibility rule for any number.

    Conjecture : Picking some particular multiple of N, and writing it in the form 10X + Y, the linear combination aX + bY that is equal to 0 or N determines the pair of coefficients (a,b) such that for any multiple of N expressed likewise, as some 10x + y, the linear combination ax + by will alway give a number that is divisible by N.

    Proof :

    Let the number for which we want to establish a divisibility rule be N.
    Let's choose some multiple of N, say mN, where m > 0.
    Write mN = 10X + Y, and let aX + bY = {0,N}
    Then for any integer k , such that kN = 10x + y, we must show that ax + by is divisible by N.

    So, we have :

    [tex]10X + Y \equiv 0 ~~(mod~N) ~~~~(1)[/tex]

    [tex]aX + bY \equiv 0~~(mod~N) ~~~~(2)[/tex]

    [tex]10x + y \equiv 0 ~~(mod~N)~~~~(3) [/tex]

    and we want to show that

    [tex]ax + by \equiv 0 ~~(mod~N) [/tex]

    From (1) :

    [tex]10X \equiv -Y => 10bX \equiv -bY ~~(mod~N)~~~~(4)[/tex]

    From (2) :

    [tex]aX \equiv -bY ~~(mod~N)~~~~(5)[/tex]

    From (4) and (5) :

    [tex](10b - a)X \equiv 0 => a \equiv 10b~~(mod~N)~, X~not~div~by~N~~~~(6) [/tex]

    Also, from (3) :

    [tex]10x \equiv -y~~(mod~N)~~~~(7) [/tex]

    Multiplying (6) and (7) :

    [tex]10ax \equiv -10by => ax \equiv -by ~~(mod~N)~, 10~not~div~by~N~~~~(8) [/tex]

    (8) is nothing but

    [tex]ax + by \equiv 0 ~~(mod~N) [/tex]

    which is the required result.

    QED



    What does all this mean ?

    Let's take the case of 7. Pick a simple multiple of 7 to determine the coefficients...say 14. We want to find (a,b) so that 1a + 4b = 0 or 7. With a=4, b=-1, you get 0. Better still, a=-1, b=2, gives 7. So, if any multiple of 7 is written as 10x + y, the combination -x + 2y will always be divisible by 7. For instance consider 518, where x=51, y=8. |2y-x| = |16-51| = 35, which is divisible by 7.

    For the case of 13, 3(1) - 1(3) = 0, so for all multiples of 13, |3x - y| will be divisible by 13.
     
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