20170628, 13:16  #1 
"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 
20170628, 13:46  #2 
May 2004
New York City
3·1,409 Posts 
Just noticed who's at the top of the list for June's solutions. Congratulations!

20170628, 13:54  #3 
"Robert Gerbicz"
Oct 2005
Hungary
2·13·53 Posts 

20170701, 04:11  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}·5·227 Posts 
An interesting one... but it has boatloads of solutions! (even if you arbitrarily fix O=0, I=1)

20170807, 20:34  #5 
"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] ? For given z where we see the same sin value, up to sign? It is easy: if sin(x)=+sin(z) then x={z,180z,180+z,360z}+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 nth 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 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 nth digit! Because if this doesn't give a solution, then there is no solution for the July puzzle. 
20170808, 20:23  #6 
"Robert Gerbicz"
Oct 2005
Hungary
2542_{8} 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 Code:
sin(30)=0.49999999999 
20170808, 22:58  #7  
Feb 2017
Nowhere
3326_{10} Posts 
Quote:
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 complexconjugate 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:
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 complexconjugate 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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