Let be k the smallest positive integer such that the binomial coefficient (109k) is less than the binomial coefficient(109+1k−1). Let a be the first (from the left) digit of and let b be the second (from the left) digit of k. What is the value of 10a+b?
+9675 (109k)<(109+1k−1)(109)!k!(109−k)!<(109+1)!(k−1)!((109+1)−(k−1))!1k(109−k)!<109+1(109−k+2)!(109−k+1)(109−k+2)<(109+1)k1500000002−√1250000003000000002<k<1500000002+√1250000003000000002
Smallest positive integer k that satisfies the inequality is 381966012. (Used some computational intelligence :P)
10a + b = 38. :)
