Question 11

The number $$5^{(6^7)}$$ is written on the board in base 10. Gia takes two of the digits at a time, erases them but appends the sum of those digits at the end. She repeats this till she ends up with one digit on the board. What is the digit that remains on the board?

Solution

Replacing two digits by their sum preserves the total digit sum modulo $$9$$. The number $$5^{6^7}$$ has remainder $$1$$ modulo $$9$$ because the powers of $$5$$ repeat with period $$6$$ modulo $$9$$ and $$6^7$$ is divisible by $$6$$. Therefore the final one digit is $$1$$.

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The number $$5^{(6^7)}$$ is written on the board in base 10. Gia takes two of the digits at a time, erases them but appends the sum of those digits at the end. She repeats this till she ends up with one digit on the board. What is the digit that remains on the board?

Solution

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