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Solution to puzzle 27: 1000th digit

What is the 1000th digit to the right of the decimal point in the decimal representation of (1 + root 2)3000 ?


Consider a(n) = (1+root 2)^n + (1-root 2)^n.

Expanding both terms using the binomial theorem, notice that the odd powers cancel, while the coefficients of even powers are all integers, and therefore an is an integer.

Then, |1 − root 2| < 1, and so (1 − root 2)n  tends to zero as n tends to infinity.
Using logarithms and/or a calculator, we find that 10−1149 < (1 − root 2)3000 < 10−1148.

Therefore (1 + root 2)3000 has 1148 nines to the right of the decimal point, and so the 1000th such digit is a 9.


Remarks

Note that a large odd exponent would generate a string of zeroes rather than nines.

As a generalization, note that (a + broot r)n + (a − broot r)n is an integer for any positive integers a, b, and r.  (Ignore the trivial case where r is a perfect square.)
Therefore as n tends to infinity, (a + broot r)n will tend to an integer if |a − broot r| < 1.


Additional puzzle

Find the first digit before and after the decimal point in (root 2 + root 3)3000.


Source: Inspired by Binet's formula for Lucas numbers

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