The number is too damm high

What is the second least signifficant digit of ( 10 + 90 ) 1000000000 \lfloor(10 + \sqrt{90}) ^{1000000000}\rfloor ?


The answer is 9.

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1 solution

Leonel Castillo
May 29, 2018

Let A = ( 10 + 90 ) k A = (10 + \sqrt{90})^k where k k is the exponent given in the problem. The only important properties of k k are that it is big and it is even. We may write A = i = 0 k ( k i ) 90 i 1 0 k i A = \sum_{i=0}^k \binom{k}{i} \sqrt{90}^i 10^{k-i} . Let's also define B = ( 10 90 ) k = i = 0 k ( k i ) ( 1 ) i 90 i 1 0 k i B = (10 - \sqrt{90})^k = \sum_{i=0}^k \binom{k}{i} (-1)^i \sqrt{90}^i 10^{k-i} . Notice that A + B = 2 i = 0 k 2 ( k 2 i ) 90 2 i 1 0 k 2 i = 2 i = 0 k 2 ( k 2 i ) 9 0 i 1 0 k 2 i A+B = 2\sum_{i=0}^{\frac{k}{2}} \binom{k}{2i} \sqrt{90}^{2i} 10^{k-2i} = 2 \sum_{i=0}^{\frac{k}{2}} \binom{k}{2i} 90^i 10^{k-2i} . What is important about this number is that it is an integer. And thus A = A + B B = A + B + B \lfloor A \rfloor = \lfloor A + B - B \rfloor = A + B + \lfloor - B \rfloor . Because B B is a positive number less than 1 we already know that B = 1 \lfloor -B \rfloor = -1 so the previous expression is just A + B 1 A + B - 1 .

To find the second digit just compute A + B 1 1 99 m o d 100 A + B - 1 \equiv -1 \equiv 99 \mod 100 . The answer is 9.

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