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Let be k the smallest positive integer such that the binomial coefficient $$\binom{10^9}{k}$$ is less than the binomial coefficient$$\binom{10^9 + 1}{k - 1}$$. 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?

Feb 4, 2019

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$$\displaystyle \binom {10^9}{k} < \binom{10^9+1}{k-1}\\ \dfrac{(10^9)!}{k!(10^9-k)!}<\dfrac{(10^9+1)!}{(k-1)!((10^9+1)-(k-1))!}\\ \dfrac{1}{k(10^9-k)!} < \dfrac{10^9+1}{(10^9-k+2)!}\\ (10^9-k+1)(10^9-k+2) < (10^9+1)k\\ 1500000002-\sqrt{1250000003000000002} < k < 1500000002 + \sqrt{1250000003000000002}$$

Smallest positive integer k that satisfies the inequality is 381966012. (Used some computational intelligence :P)

10a + b = 38. :)

Feb 4, 2019