# Solution to puzzle 27: 1000th digit

What is the 1000th digit to the right of the decimal point in the decimal representation of (1 + )^{3000} ?

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

Then, |1 − | < 1, and so (1 − )^{n} tends to zero as n tends to infinity.

Using logarithms and/or a calculator, we find that 10^{−1149} < (1 − )^{3000} < 10^{−1148}.

Therefore (1 + )^{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 + b)^{n} + (a − b)^{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 + b)^{n} will tend to an integer if |a − b| < 1.

## Additional puzzle

Find the first digit before and after the decimal point in ( + )^{3000}.

Source: Inspired by Binet's formula for Lucas numbers

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