According to Euler's Theorem, if gcd(a,n) = 1, then a(n) 1. (mod n). The number of relatively prime integers in the range of 1, 2,..., n-1 is represented by the Euler totient function.
The probability that mn has a units digit of 1 = 2/5
By Euler's Theorem, we have that [tex]$a^{4} \equiv 1 \pmod {10}$[/tex] if [tex]$\gcd(a,10)=1$[/tex] Hence[tex]$m=11,13,17,19$,[/tex][tex]$n=2000,2004,2008,2012,2016$[/tex] work.
Also note that [tex]$11^{\text{any positive integer}}\equiv 1 \pmod {10}$[/tex] because [tex]$11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1$[/tex], and the latter[tex]$\pmod {10}$[/tex]is clearly [tex]$m=11$[/tex], [tex]$n=1999,2001,2002,2003,2005,...,2018$[/tex] work (not counting multiples of 4 as we would be double counting if we did).
We can also note that [tex]$19^{2a}\equiv 1 \pmod {10}$[/tex] because[tex]$19^{2a}=361^{a}$[/tex], and by the same logic as why [tex]$11^{\text{any positive integer}}\equiv 1 \pmod {10}$[/tex]we are done. Hence [tex]$n=2002, 2006, 2010, 2014, 2018$[/tex] and [tex]$m=19$[/tex] work (not counting any of the aforementioned cases as that would be double counting).
We cannot make any more observations that add more[tex]$m^n$[/tex] with unit digit [tex]$1$[/tex], hence the number of [tex]$m^n$[/tex] that have units digit one is [tex]$4\cdot 5+1\cdot 15+1\cdot 5=40$.[/tex] And the total number of combinations of an element of the set of all [tex]$m$[/tex]and an element of the set of all n is [tex]$5\cdot 20=100$[/tex]. Hence the desired probability is [tex]$\frac{40}{100}=\frac{2}{5}$[/tex].
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