Solving nⁿ=20880467999847912034355032910567
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Here's a nice equation: what is n if nⁿ=20880467999847912034355032910567
provided that n is an integer.

How do we approach it?

1. Count the digits:

12345678901234567890123456789012
20880467999847912034355032910567

2. Find where the solution lies:

The number of digits in 20880467999847912034355032910567 is 32.
Thus, n cannot be a sinlge-digit number because
9⁹<10⁹

n has two digits.
And 32=16 × 2
And because 100¹⁵=10³⁰, whose lenght is only 31, thus

  n>15

And the smallest two digit number is 10, and 10³² is of length 33, thus

  n<32

3. Elimination:

n is an odd number whose unit digits cannot be 1,5 or 9. That leaves
us with 17,23 and 27
The last two digits of nⁿ are 67, which means that nⁿ leaves a remainder of
3 when divided by 4, which means that:

  n ≡ 3 mod 4

which leves us with 23 and 27, but

27²⁷ ≡ 7³ mod 10 ≡ 3 mod 10

That leaves us with

  n = 23

our solution.