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## El Cheapo Number Game

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)

#2. Multiply this by 7

#3. Write this number in the form 10X + Y, where $0 \leq Y \leq 9$ (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.

Debunk the magic (if it worked ).
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 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
 Recognitions: Gold Member Science Advisor Staff Emeritus Thanks for pointing that out. I forgot to put in a modulus sign (absolute value).

## El Cheapo Number Game

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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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 ?
 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.

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 Recognitions: Gold Member Science Advisor Staff Emeritus 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 : $$10X + Y \equiv 0 ~~(mod~N) ~~~~(1)$$ $$aX + bY \equiv 0~~(mod~N) ~~~~(2)$$ $$10x + y \equiv 0 ~~(mod~N)~~~~(3)$$ and we want to show that $$ax + by \equiv 0 ~~(mod~N)$$ From (1) : $$10X \equiv -Y => 10bX \equiv -bY ~~(mod~N)~~~~(4)$$ From (2) : $$aX \equiv -bY ~~(mod~N)~~~~(5)$$ From (4) and (5) : $$(10b - a)X \equiv 0 => a \equiv 10b~~(mod~N)~, X~not~div~by~N~~~~(6)$$ Also, from (3) : $$10x \equiv -y~~(mod~N)~~~~(7)$$ Multiplying (6) and (7) : $$10ax \equiv -10by => ax \equiv -by ~~(mod~N)~, 10~not~div~by~N~~~~(8)$$ (8) is nothing but $$ax + by \equiv 0 ~~(mod~N)$$ 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.