IBM Research | Ponder This | November 2012 solutions
Skip to main content

November 2012

<<October November December>>


The problem was originally proposed almost a century ago (see Pólya, G., "Zahlentheoretisches und wahrscheinlichkeitstheoretisches über die sichtweite im walde," Arch. Math. Phys., 27 (1918): 135-142).

The answer is 1/sqrt(d), where d is the smallest integer satisfying d=a^2+b^2, where a is co-prime to b, and d>=R^2.

Clearly, d<=R^2+1 by choosing a=R and b=1; and d can be R^2 if and only if all the prime factors of R are 1 modulo 4.

In our case, R=9801=3^4*11^2, so the critical radius is 1/sqrt(R^2+1) which is 0.000102030404529629...


If you have any problems you think we might enjoy, please send them in. All replies should be sent to: webmster@us.ibm.com