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 2017-06-28, 13:16 #1 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 2×13×53 Posts July 2017 The new Ibm puzzle is out: https://www.research.ibm.com/haifa/p.../July2017.html
 2017-06-28, 13:46 #2 davar55     May 2004 New York City 3·1,409 Posts Just noticed who's at the top of the list for June's solutions. Congratulations!
2017-06-28, 13:54   #3
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

2·13·53 Posts

Quote:
 Originally Posted by davar55 Just noticed who's at the top of the list for June's solutions. Congratulations!
Thanks! I've a really good year: 5 first place out of 6.

 2017-07-01, 04:11 #4 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23·5·227 Posts An interesting one... but it has boatloads of solutions! (even if you arbitrarily fix O=0, I=1)
 2017-08-07, 20:34 #5 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 2·13·53 Posts The official solution is at: https://www.research.ibm.com/haifa/p.../July2017.html actually you can do it easier: it isn't discussed, but here it is a short code to find those angles (in degrees), where sin(u)*sin(v) has 10 zeros after the tenth decimal digit, we can assume that we are in the first quadrant, because up to sign we see all possible sin values there (and we can also assume that u,v != k*90, otherwise sin(u)=0,+-1, that isn't good). Up to symmetry for solution: Code: for(i=1,89,for(j=i,89,a=sin(i/180*Pi)*sin(j/180*Pi);\ x=round(a*10^20);if(x%(10^10)==0,print([i,j,a])))) [15, 75, 0.25000000000000000000000000000000000000] [18, 54, 0.25000000000000000000000000000000000000] [30, 30, 0.25000000000000000000000000000000000000] [45, 45, 0.50000000000000000000000000000000000000] [60, 60, 0.75000000000000000000000000000000000000] ? (we really need round or eps to avoid precision problem) For given z where we see the same sin value, up to sign? It is easy: if sin(x)=+-sin(z) then x={z,180-z,180+z,360-z}+360*k for integer k. so for integer x,y it is true that x==z,-z mod 10, but if z==0,5 mod 10, then x==z mod 10 So excluding the 2nd case (u=18;v=54) all of them is impossible: bacause in those cases x1==x2=={0,5} mod 10 R==S mod 10, but these are digits, hence R=S, what is impossible! So you don't need the "sin(15), sin(30), sin(45), sin(60), and sin(75) have an 863172th digit of "0";" part. These can't give a solution... Furthermore it is easy to find the n-th decimal digit (after the decimal point) using only integer arithmetic of sin(18)=(sqrt(5)-1)/4, for n>2 the digit is: Code: f(n)=(sqrtint(5*10^(2*n))\4)%10 ps. found this f(n) trick after I've sent my solution, and the first part of the above argument wasn't needed if you give the letters value, so a solution. ps2. if you accept that there is really a solution for the puzzle, then it can be only the sin(18)*sin(54)=1/4 (up to signs/symmetry and mod 180 for degree), so you don't need to check the n-th digit! Because if this doesn't give a solution, then there is no solution for the July puzzle.
 2017-08-08, 20:23 #6 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 25428 Posts There is some issue with the current wording: a rational number can have multiple decimal representation (at most two), and this has a big role here: Code: sin(18)*sin(54)=0.25=0.2499999999999 so using the other form the official solution has not even a single digit zero after the decimal point. And this is also a problem for sin(30): Code: sin(30)=0.49999999999 so the 863172th digit of this is not zero, only the other form has zero. (here my above proof is still correct, there is no solution in the sin(30)*sin(30) case).
2017-08-08, 22:58   #7
Dr Sardonicus

Feb 2017
Nowhere

332610 Posts

Quote:
 Originally Posted by R. Gerbicz There is some issue with the current wording: a rational number can have multiple decimal representation (at most two), and this has a big role here: Code: sin(18)*sin(54)=0.25=0.2499999999999 so using the other form the official solution has not even a single digit zero after the decimal point. And this is also a problem for sin(30): Code: sin(30)=0.49999999999 so the 863172th digit of this is not zero, only the other form has zero. (here my above proof is still correct, there is no solution in the sin(30)*sin(30) case).
Setting aside the wording issue, I note that "solutions" based on sin(30)*sin(30) would violate the digits condition; R and S would both have to be 0.
Some may find the "theoretical" approach I took amusing.

First, I rejected out of hand the idea that any irrational value of sin(A)*sin(B) (with A = PONDER*Pi/180 and B = THIS*Pi/180, PONDER and THIS integers), would have over 20% of the first million digits equal to 0. Any product sin(A)*sin(B) I was going to find would have to be rational.

(My hat's off to Andreas Stiller, who actually checked!)

I then fell back on a trig identity to rewrite the product:

2*sin(A)*sin(B) = cos(A - B) - cos(A + B), so that

4*sin(A)*sin(B) = 2*cos(A - B) - 2*cos(A + B).

Now here, cos(A+B) and cos(A - B) are the real parts of roots of unity, so 2*cos(A - B) and 2*cos(A + B) are each sums of two complex-conjugate roots of unity, and are therefore algebraic integers.

Thus, if 4*sin(A)*sin(B) is rational, it is a rational integer. Thus, if sin(A)*sin(B) is a fraction, its denominator divides 4. The product cannot be 0, because of the condition
Quote:
 the 863,172nd digit after the decimal point of sin(PONDER) and of sin(THIS) are not zeros.
Now, the only possible rational values for 2*cos(A - B) and 2*cos(A + B) are -2, -1, 0, 1, and 2. These are easily ruled out by the condition that R and S be different.

Thus, any values 2*cos(A - B) and 2*cos(A + B) which lead to solutions must be irrational. If 2*cos(A - B) - 2*cos(A + B) = n, a rational integer, they have the same degree d > 1.

Now if the sum of the algebraic conjugates of 2*cos(A + B) is S, then the sum of the algebraic conjugates of 2*cos(A - B) is S + n*d. Since d > 1, and the sine product cannot be 0, the difference of the two sums must have absolute value at least 2.

But in each case, the sum is equal to the sum of the primitive roots of unity of some order; and this sum can only be 1, -1, or 0. Therefore, the difference has absolute value 2, d is 2, and n is 1 or -1.

The only possible minimum polynomials for 2*cos(A - B) and 2*cos(A + B) are x^2 + x - 1 and x^2 - x - 1 (for sums of two complex-conjugate fifth and tenth roots of unity, respectively). These lead to solutions.

(Incidentally, I didn't bother checking that the 863,172nd digits of the sines wasn't 0 until after I'd sent in my solution; I figured that if they were, the problem had no solution. When I did check, it was by finding a file online with the first million digits of the "golden ratio" phi. The 3 digits ending in the 863,172nd were 138, forcing the 863,172nd digit of phi/2 to be 9.)

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