FreddieCodes - 1 year ago 40

Ruby Question

Here's my code:

`def last_digit(n1, n2)`

array = (n1.to_i ** n2.to_i).to_s.split("")

array[-1].to_i

end

TEST: The last decimal digit of (2^200)^(2^300), which has over 10^92 decimal digits, is 6

I'm trying to return the last digit of a last number and I'm sure this correct but when I run tests 2 return as failing.

I think it's due to the numbers being too large, how do I get this code to remain accurate no matter how large it gets.

And also how do I deal with NaN, I've searched and struggled to find anything useful.

Thanks for your help.

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Answer Source

There's an effective algorithm which assumes that only the last digit of a number being powered matters. Please, try it out on your tests and feel free to correct any flaw in this implementation that you'll find by running them

```
def digit_of_power(digit, n)
digit = digit % 10
case digit
when 0, 1, 5, 6 then digit
else
digit_of_square = digit * digit
if n.even?
digit_of_power(digit_of_square, n / 2)
else
digit * digit_of_power(digit_of_square, (n - 1) / 2) % 10
end
end
end
```

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